# Math Linear Algebra II. 1. Inner Products and Norms

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 Math Linear Algebra II Notes 1. Inner Products and Norms One knows from a basic introduction to vectors in R n Math 254 at OSU) that the length of a vector x = x 1 x 2... x n ) T R n, denoted x, or x, is given by the formula x = x x x 2 n. We will call this this function : R n [0, ) the standard) norm on R n. Recall the familiar dot product on R n : Let x = x 1 x 2... x n ) T, y = y 1 y 2... y n ) T R n. Then n x y = x i y i. We will call the dot product on R n the standard) inner product on R n, and we will use the fancier notation x, y) for x y. Formally, we can regard the inner product as a function, ) : R n R n R. An easy and important consequence is that x, x) = x 2, equivalently, x = x, x). We also have another expression for this inner product: For all x, y R n, x, y) = y T x same as x T y).

2 Now let s explore the standard) norm and the standard) inner product on C n : Recall that for z C, z = a + bi, and z 2 = a 2 + b 2 = z z really just the R 2 standard) norm!) Now let z = z 1 z 2... z n ) T C n with z j = a j + b j i for j = 1, 2,..., n. It is natural to define the standard) norm on C n by n n z = z j 2 = a 2 j + b2 j ) for z Cn. j=1 j=1 Furthermore, with w = w 1 w 2... w n ) T C n, we define the standard) inner product on C n by z, w) = n z i w i. Again, an easy and important consequence is that z, z) = z 2, equivalently, z = z, z). Let A M mn have entries in C. Let A = Ā)T where Ā determined by applying the conjugate operation entry-wise. This combination conjugate transpose) is the hermitian adjoint or just the adjoint of A. For w, w = w) T = w 1 w 2... w n ) and thus z, w) = w z.

3 Now let s get more abstract. We will let F denote either R or C. Let V be an arbitrary vector space over F. An inner product on V is a function, ) : V V F, that takes in two vectors u, v V and outputs a scalar u, v) F, satisfying these four fundamental properties: 1) Positivity aka non-negativity): For all v V, v, v) R and v, v) 0. 2) Definiteness aka non-degeneracy): Let v V. Then v, v) = 0 iff v = 0. 3) Linearity: For all c F, for all u, v, w V, u + cv, w) = u, w) + cv, w). 4) Conjugate symmetry: For all u, v V, u, v) = v, u) just symmetry if F = R.) If V is a vector space over F and, ) is an inner product on V then we call V and inner product space over F. Examples: In all of these, if F = R then there is no need for the conjugation.) 1) Let V = F n. We have already covered standard aka Euclidean) inner product on V given by For all u, v V, u, v) = v u. ELFY) Show that this function does indeed satisfy the four fundamental properties of an inner product... 2) Let V = P n, polynomials of degree at most n over F. Let the interval I = [a, b]. Then for fx), gx) P n define fx), gx)) I = b a fx)gx)dx. ELFY) Show that this function does indeed satisfy the four fundamental properties of an inner product...

4 3) Let V = M m,n over F). For A, B M mn define A, B) = traceb A). ELFY) Show that this function does indeed satisfy the four fundamental properties of an inner product... Note: traceb A) = n B A) ii = n b 1i a 1i + b 2i a 2i + + b mi a mi ). Properties of inner products: From here until otherwise indicated page 12), let V be an inner product space over F). ELFY) Show that For all c F, for all u, v, w V, u, v + cw) = u, v) + cu, w). Hint: Use conjugate symmetry, linearity, and then conjugate symmetry again! Proposition: Let x V. Then, x = 0 iff for all y V, x, y) = 0. Proof : The forward direction is trivial: For all y V, 0, y) = 0y, y) = 0y, y) = 0... also y, 0) = y, 0y) = 0y, y) = 0). Now assume x V satisfies for all y V, x, y) = 0. Then in particular, x, x) = 0. Using the definite property of an inner product we get x = 0 ELFY) Use the proposition above to prove that for all x, y V, x = y iff for all z V, x, z) = y, z).

5 Proposition: Let T, S LV ). Suppose that for all x, y V, T x, y) = Sx, y). Then T = S. Proof : Let T, S LV ). *) Assume for any x, y V that T x, y) = Sx, y). Assume T S. Then there exists x V such that T x Sx. However, T x, y) = Sx, y) for all y V by assumption *). Using the exercise left for you on the previous page, T x = Sx, which is a contradiction. So T = S An inner product on V induces a norm on V corresponding to the inner product, which is a function : V R given by Basic norm property: v = v, v) for v V. cv = cv, cv) = c cv, v) = c 2 v, v) = c v. Also, we call two vectors u, v V orthogonal if u, v) = 0 by conjugate symmetry, v, u) = 0 would also hold). ELFY) if u, v are orthogonal, then u + v 2 = u 2 + v 2 Pythagorean Theorem).

6 Orthogonal decomposition: Let u, v V. Assume v 0. Then v, v) 0. Let c F. Clearly, u = cv + u cv). Let s find c such that v, u cv) are orthogonal: u cv, v) = 0 u, v) cv, v) = 0 u, v) = cv, v) c = u, v) v, v) = u, v) v 2. We will call proj v u) = u,v) v 2 v the orthogonal projection of u onto v. We will call u = proj v u) + u proj v u)) the orthogonal decomposition of u along v Consider that ) u, v) ) proj v u) = v = v 2 u, v) v. Notice that since u proj v u), v) = 0 it follows that u proj v u), proj v u)) = u proj v u), u, v) u, v) v) = v 2 v u proj vu), v) = 0. 2 Thus if u = proj v u) + u proj v u)) is the orthogonal decomposition of u along v, then the vector terms satisfy the Pythagorean Theorem: ) u 2 = proj v u) 2 + u proj v u) 2. Then ) and ) implies u proj v u) 2 = u 2 u, v) 2 v 2.

7 You could of course calculate this directly: u proj v u) 2 = u proj v u), u proj v u)) = u, u proj v u)) proj v u), u proj v u)) = u, u) u, proj v u)) proj v u), u) + proj v u), proj v u)) = u 2 + u, v) 2 v 2 u, proj v u)) u, proj v u)) = u 2 + = u 2 + u, v) 2 v 2 u, u, v) 2 v 2 u, v) v u, v) u, v) v) u, v 2 v v) 2 u, v) u, v) u, v) 2 v 2 = u 2 + u, v) 2 v 2 u, v) 2 v 2 u, v) 2 v 2 = u 2 u, v) 2 v 2. In the last step we used the fact that conjugation does nothing to a real number: u, v) 2 v 2 = u, v) 2 v 2.

8 The following propositions is of great importance. Proposition Cauchy-Schwarz Inequality): If u, v V then u, v) u v. Proof : Let u, v V. If v = 0 then the inequality holds as both sides are clearly 0. So we assume that v 0. Consider the orthogonal decomposition u = proj v u) + w where w = u proj v u)). 0 w 2 = u 2 u, v) 2 v 2 u, v) 2 v 2 u, v) v u 2 u u, v) u v

9 Proposition Triangle Inequality): If u, v V then u + v u + v. Proof : Let u, v V. Consider that for all z = a + bi C, Rea + bi) = a a = a 2 a 2 + b 2 = a + bi. Then u + v 2 = u + v, u + v) = u, u) + v, v) + u, v) + v, u) = u 2 + v 2 + u, v) + u, v) = u 2 + v 2 + 2Re[u, v)] u 2 + v u, v) u 2 + v u v = u + v ) 2. Hence, u + v u + v

10 These are the so-called polarization identities: Proposition: For all x, y V, 1) x, y) = 1 4 x + y 2 x y 2) if F = R. 2) x, y) = 1 4 x + y 2 x y 2 + i x + iy 2 i x iy 2) if F = C. Proof : Let x, y V. Assume F = R. Then conjugation does nothing and x, y) = y, x). Consider that 1 x + y 2 x y 2) = 1 x + y, x + y) x y, x y)) = 4 4 = 1 x, x) + 2x, y) + y, y) [x, x) 2x, y) + y, y)]) = x, y). 4 Assume F = C. Then z = 1 4 x + y 2 x y 2 + i x + iy 2 i x iy 2) = = 1 x + y, x + y) x y, x y) + ix + iy, x + iy) ix iy, x iy)) 4 Let s split z into two parts: and c 1 = 1 x + y, x + y) x y, x y)), 4 c 2 = 1 ix + iy, x + iy) ix iy, x iy)). 4 Clearly z = c 1 + c 2.

11 c 1 = 1 x, x) + x, y) + y, x) + y, y) [x, x) x, y) y, x) + y, y)]) 4 c 1 = 1 4 2x, y) + 2y, x)) = 1 x, y) + y, x)). 2 c 2 = i x, x) + x, iy) + iy, x) + iy, iy) [x, x) x, iy) iy, x) + iy, iy)]) 4 c 2 = i 4 2x, iy) + 2iy, x)) = i 2 [ i)x, y) + iy, x)] = 1 x, y) y, x)). 2 Finally, z = c 1 + c 2 = x, y) Proposition Parallelogram Identity): For all x, y V, x + y 2 + x y 2 = 2 x 2 + y 2). I leave the proof of this one to you...it s pretty easy.

12 Now let V be a vector space over F, but we no longer assume that there is an inner product on V. The function n : V R is a norm on V if it satisfies the following properties: 1) Homogeneity: For all v V and for all c F, ncv) = c nv). 2) Triangle Inequality: For all u, v V, nu + v) nu) + nv). 3) Positivity aka non-negativity): For all v V, nv) 0. 4) Definiteness aka non-degeneracy): Let v V. Then nv) = 0 iff v = 0. Of course, the norm : V R induced from an inner product space V satisfies these properties: We have already verified that 1) and 2) hold. Properties 3) and 4) are direct consequences of the positivity and definiteness of the inner product. So all inner product spaces are normed spaces, via the induced norm v = v, v). However, there are normed spaces where the norm does not correspond to an inner product. When V be a normed vector space, with norm n, we will still use the notation nv) = v.

13 Let V = R n or V = C n. p : V R given by It turns out that for any p, 1 p <, following function x p = n x i p ) 1 p, is a norm called the p-norm. Of course the 2-norm aka Euclidean norm) is the norm induced by the standard inner product on V. We can also define a norm called the -norm: x = max{ x i : i = 1, 2,..., n}. For p and it is easy to check ELFY) that they satisfy defining properties 1), 3), and 4) of a norm. The triangle inequality is a different story. Of course we already have proven the triangle inequality for 2. The triangle inequality for p isn t too hard to check ELFY) when p = 1 and p =. For arbitrary values of p it is fairly involved to prove the triangle inequality, but it does hold, and is called Minkowski s Inequality. We won t do that here... It turns out that p for p 2 cannot be induced by an inner product they don t satisfy the parallelogram inequality)! The proof is beyond the scope of this course because of the reverse implication), but the following proposition characterizes norms induced from inner products: Proposition: Let V be a normed space with a norm. The norm is induced from an inner product, ) on V iff satisfies the parallelogram inequality, that is, for all x, y V, x + y 2 + x y 2 = 2 x 2 + y 2). Suggested Homework: 1.1, 1.2, 1.3, 1.5, 1.7, and 1.9.

14 2. Orthogonal and Orthonormal Bases Recall two vectors u, v in an inner product space V are orthogonal if u, v) = 0. We will use the notation u v. A vector v V is orthogonal to a subspace U of V if for all u U, u v. We use the notation v U. Two subspaces U 1, U 2 are orthogonal subspaces if for all u 1 U 1, u 1 is orthogonal to the subspace U 2. Furthermore, we can define for any subspace U of V, the orthogonal complement to U in V ): U = {v V v U}. Also u is called a unit vector if u, u) = 1 and thus u = 1). The vectors v 1, v 2,..., v k are called an orthogonal system if they are pairwise orthogonal, and an orthonormal system if they are an orthogonal system of unit vectors. Proposition: Any orthogonal system of non-zero vectors is linearly independent. Proof : Let v 1, v 2,..., v k be an orthogonal system of non-zero vectors. Suppose c i v i = 0 for c i F, i = 1, 2,..., k. Then for any j = 1, 2,..., k we get ) c i v i, v j = c i v i, v j ) = c j v j 2 = 0. For all j = 1, 2,..., k, v j 2 0 as v j 0. Therefore, c 1 = c 2 =... = c k = 0, which completes the proof that the orthogonal system is linearly independent

15 Proposition General Pythagorean Theorem): Let v 1, v 2,..., v k be an orthogonal system. Then for all c 1, c 2,..., c k F, 2 c i v i = c i 2 v i 2. Proof : Let v 1, v 2,..., v k be an orthogonal. Let c i F for i = 1, 2,..., k. 2 c i v i = c i v i, ) c j v j = j=1 c i c j v i, v j ) = j=1 c i c i v i, v i ) = c i 2 v i 2 A basis B = {v 1, v 2,..., v n } of V is an orthogonal basis orthonormal basis) if B is an orthogonal system orthonormal system). Due to a previous proposition, if n = dim V, then any orthogonal orthonormal) system of n non-zero vectors forms a orthogonal orthonormal) basis. Let B = {v 1, v 2,..., v n } be an orthogonal basis of V. Suppose x V. Usually finding [x] B = c 1 c 2 c n ) T involves solving a linear system, but it is easier when B is an orthogonal basis. For any j = 1, 2,..., n, n ) x, v j ) = c i v i, v j = n c i v i, v j ) = c j v j, v j ) = c j v j 2 c j = x, v j) v j 2. This should look familiar! Recall that proj v u) = u,v) v 2 v.) Furthermore, if B is an orthonormal basis, this result reduces to c j = x, v j ).

16 When B = {v 1, v 2,..., v n } is an orthonormal basis, then for any v V, the formula v = n v, v i )v i, is called an orthogonal Fourier decomposition. Suggested Homework: 2.2, 2.3, 2.5, 2.6, 2.8, and 2.9.

17 3. Orthogonalization and Gram-Schmidt Let W be a subspace of V. For each v V, w W is called an orthogonal projection of v onto W if v w) W. Proposition: Let W be a subspace of V and let w be an orthogonal projection of v onto W. Then 1) For all x W, v w v x w minimizes distance from v to W ). 2) If x W and v w = v x then x = w orthogonal projections are unique). Proof : Let W be a subspace of V and let w W be an orthogonal projection of v onto W. Let x W. Set y = w x W. Then v x = v w) + y. Since v w) y as y W, by the Pythagorean Theorem we get v x 2 = v w 2 + y 2 v x v w, and v x = v w iff y = 0. Hence v x = v w iff x = w Now we can say the orthogonal projection of v onto W.) Let W be a subspace of V with an orthogonal basis B = {v 1, v 2,..., v k }. Let v V and P W v be the orthogonal projection of v onto W. Let w = v, v i ) v i 2 v i W. Then v w = v v, v i ) v i 2 v i.

18 Let x W. Then there exists scalars c 1, c 2,..., c k such that x = k c iv i. Then v w, x) = = v v, v, v i ) v i 2 v i, ) c j v j j=1 ) c i v i v, v i ) v i 2 v i, ) c j v j j=1 = = = = c i v, v i ) c i v, v i ) c i v, v i ) c i v, v i ) j=1 v, v i ) v i 2 c j v i, v j ) v, v i ) v i 2 c i v i, v i ) v, v i ) v i 2 c i v i 2 c i v, v i ) = 0. This shows that w W satisfies v w) W, and so, w is the orthogonal projection of v onto W. Hence we get the formula: v, v i ) P W v = v i v i. 2 Claim: Let W be a subspace of V. P W LV ). Let u, v V and c F. Let w W. Then P W u + cp W v W and [u + cv] [P W u + cp W v], w) = [u P W u] + c[v P W v], w) = u P W u, w) c v P W v, w) = 0. Thus P W u + cv) = P W u + cp W v showing that P W is a linear operator on V.

19 The Gram-Schmidt algorithm: Input: A linearly independent set of vectors {x 1, x 2,..., x k } spanning a subspace W of V. Output: An orthogonal system {v 1, v 2,..., v k } such that the span is W. Here are the steps: 1) Set v 1 = cx 1 for any non-zero c F usually 1). Set W 1 = span{x 1 } = span{v 1 }. ) 2) Set v 2 = cx 2 P W1 x 2 ) = c x 2 x 2,v 1 ) v 1 v 2 1 for any non-zero c F. Set W 2 = span{v 1, v 2 }. Note that W 2 = span{x 1, x 2 }. ) 3) Set v 3 = cx 3 P W2 x 3 ) = c x 3 x 3,v 1 ) v 1 v 2 1 x 3,v 2 ) v 2 v 2 2 for any non-zero c F. Set W 3 = span{v 1, v 2, v 3 }. Note that W 3 = span{x 1, x 2, x 3 }. 4) Continue in this manner until v k is determined it s not necessary to determine W k ). Suppose V = R 3. Consider the basis B = 1 2 3, 1 1, Let s apply the Gram-Schmidt procedure to turn this into an orthogonal basis: 1) Set v 1 = and W 1 = span

20 2) Consider that P W = = , = = So for convenience set v 2 = 5 4 1, and set W 2 = span 1 2, = 3) Consider that , 1 = P W = 2 1, = = So for convenience set v 3 =

21 In conclusion B = 1 2 3, 5 4, 1 1, is an orthogonal basis 1 1 such that the first i vectors of B and B have the same spans for i = 1, 2, 3. Recall if W is a subspace of V, then W = {v V v W }. Claim: Let W be a subspace of V. W orthogonal complement) is also a subspace of V. This is pretty easy to justify: First notice W is not empty as 0 W. Let u, v W and c F. Let w W. Then u + cv, w) = u, w) + cv, w) = 0. Thus u + cv W proving that W is a subspace of V. More generally, if U, W are subspaces of V then U + W = {u + w u U, w W } is called the subspace sum. ELFY) U + W is a subspace of V for any subspaces U, W. A subspace sum U + W is called a direct sum, denoted U W, if U W = {0}. Proposition: Let U, W be subspaces of V. V = U W iff each vector in v V has a unique representation v = u + w for u U, w W. Proof : Let U, W be subspaces of V. Assume V = U W. Let v V. Then there exists u U and w W such that v = u+w. Suppose v = u 1 +w 1 where u 1 U and w 1 W. Then v = u + w = u 1 + w 1 implies u u 1 = w w 1. Then u + u 1 = w w 1 U V = {0}. Hence u = u 1 and w = w 1, so we have uniqueness.

22 Now assume for all v V there exists unique vectors u U, w W such that v = u + w. We are assuming each vector in V has a unique decomposition into a vector from U plus a vector from W. Clearly V = U + W. The question is whether it s a direct sum. To yield a contradiction, suppose U W {0}. Let y U W such that y 0. Let v V. Since V = U + W there exists u U, w W such that v = u + w. Set u 1 = u + y U. Since y 0, u 1 u. Set w 1 = w y U. Since y 0, w 1 w. Then u 1 + w 1 = u + w = v, which contradicts the uniqueness of the decomposition. This completes the proof Let v V and W be a subspace at V. Then let v 1 = P W v W and v 2 = v P W v W. We have already shown that P W v W is the unique vector for which v P W v W. Therefore we have a unique decomposition of v into a vector in W plus a vector in W. Hence V = W W. ELFY) For any subspace W of V, we get W ) = W. Suggested Homework: 3.2, 3.3, 3.4, 3.5, 3.7, and 3.8.

23 4. Least Squares Solution Let A M mn and b F m. How do we determine an x F n such that Ax is as close as possible to b? If we can find x such that Ax b = 0 then Ax = b is consistent and we say x is an exact solution. In the case where we cannot find x such that Ax b = 0 then we call an x which minimizes Ax b a least squares solution to Ax = b which is inconsistent). Why least squares? Well, minimizing Ax b is equivalent to minimizing m m Ax b 2 = Ax) i b i 2 n 2 = A ij x j b i. j=1 Let W = range A = {Ax x F n }. W is a subspace of F m. In particular, the minimum value of Ax b is the distance from b to W, which is satisfied by an x which solves the equation *) Ax = P W b. Furthermore, if v 1, v 2,..., v k is an orthogonal basis for W = range A then we get an explicit formula for P W b: P W b = b, v i ) v i 2 v i. So, if we have a basis for W which we can create by taking columns of A that are pivot columns) then with Gram-Schmidt we can create an orthogonal basis and compute P W b and then solve *). That is actually a lot of work...maybe there is an easier method? Here is a nice approach: Let A = [a 1 a 2 a n ]. Ax = P W b iff b Ax) W iff b Ax) a i for i = 1, 2,..., n. Hence Ax = P W b iff b Ax), a i ) = a i b Ax) = 0 for i = 1, 2,..., n. Here is the clever part: A b Ax)) i = a i b Ax) = 0 for i = 1, 2,..., n. So Ax = P W b iff A b Ax) = 0 iff A Ax = A b.

24 So we get the following result: So the easiest way to find a least squares solution to Ax = b is to solve the so-called normal equation A Ax = A b. Also for a least squares solution x, P W b = Ax In particular, this means that a least squares solution is unique iff A A M nn is invertible. In this case P W b = AA A) 1 A b, which would hold for all b F m, so then the matrix of P W is AA A) 1 A. Proposition: Let A M mn. Ker A = KerA A). In other words, A A is invertible iff rank A = n. Proof : Let A M mn. Suppose x Ker A. Then Ax = 0 and thus A A)x = A Ax) = A 0 = 0. Hence x KerA A) and Ker A KerA A). Suppose x KerA A). Then A A)x = A Ax = 0. Thus Ax 2 = Ax, Ax) = Ax) Ax = x A )Ax = x A Ax) = x 0 = 0. This implies that Ax = 0. So x Ker A and KerA A) Ker A. This completes the proof Line of best fit: A classic application is finding a line y = mx+b to best fit some given data {x 1, y 1 ), x 2, y 2 ),..., x k, y k )}. The best solution minimizes the following sum of square differences: mx i + b y i 2.

25 Equivalently, we are looking for a least squares solution to x 1 1 x x k 1 m b ) = y 1 y 2... y k. As long as not all x-coordinates are the same, a least squares solution is unique. EXAMPLE: Suppose we have the following data points in the xy-plane: {1, 2), 3, 5), 4, 5), 6, 8), 6, 9), 7, 10)}. Find an equation for the line of best fit. So we are looking for the least squares solution to m b ) = First we create the normal equation: ) m b ) = ) ) m b ) = ) m b ) = ) ) =

26 Hence the line of best fit is... y = x General approach to curve-fitting or surface-fitting: 1) Find equations your data should fit if there was an exact solution. The equations must be linear with respect to the parameters.) 2) Write these equations as a linear system in a matrix form inconsistent unless there is an exact solution). 3) Find the least squares solution left-multiply both sides by the conjugate transpose, aka adjoint, of the coefficient matrix and solve). Suggested Homework: 4.2, 4.3, and 4.4.

27 5. Adjoint of a Linear Transformation Let V = F n. Let A M mn with entries in F R or C). Recall the adjoint of A is A = A T. Proposition: Let A M mn. For all x, y V, Ax, y) = x, A y). Proof : Let A M mn and x, y V. Then Ax, y) = y, Ax) = Ax) y = x A )y = x A y) = A y, x) = x, A y) This proposition gives rise to the notion of an adjoint operator. Now let V, W be finite dimensional) inner product spaces and T LV, W ). S LW, V ) is called an adjoint operator if T x, y) = x, Sy) for all x V and y W. We will use the notation T for S. Does it exist? Yes. Let T LV, W ). By having B, C be orthonormal bases for V, W respectively, define T LW, V ) by having [T ] BC = [T ] CB ). Let s justify that it is indeed adjoint: Let x V and y W. Let B = {v 1, v 2,..., v n } and C = {w 1, w 2,..., w m } be orthonormal bases for V, W respectively. Then T v i = c 1i w 1 + c 2i w c mi w m and y = d 1 w 1 + d 2 w d m w m, where c ij, d k are scalars for i = 1, 2,..., m, j = 1, 2,..., n, and k = 1, 2,..., m. Then T v i, y) = m k=1 c kid k = [T ] CB [v i ] B, [y] C ) = [v i ] B, [T ] CB ) [y] C ) for i = 1, 2,..., n. By linearity, we get T x, y) = [T ] CB [x] B, [y] C ) = [x] B, [T ] CB ) [y] C ). Also T w i = a 1i v 1 + a 2i v a ni v n and x = b 1 v 1 + b 2 v b n v n, where a ij, b k are scalars for i = 1, 2,..., n, j = 1, 2,..., m, and k = 1, 2,..., n. Then x, T w i ) = n k=1 a kib k = [x] B, [T ] BC [w i ] C ) = [x] B, [T ] CB ) [w i ] C ) for i = 1, 2,..., m. By linearity, we get x, T y) = [x] B, [T ] CB ) [y] C ) = T x, y). So T is indeed an adjoint operator to T. Is it unique? Suppose S is another adjoint operator. Then for all x V and y W, T x, y) = x, T y) = x, Sy) implies that for all y W, T = Sy. Thus S = T. So the adjoint operator is unique.

28 Proposition: Let T LV, W ) where V, W are inner product spaces. Then 1) Ker T = range T ). 2) Ker T = range T ). 3) range T = Ker T ). 4) range T = Ker T ). Proof : First recall that for any subspace U, U ) = U. Thus 1) and 3) are equivalent as well as 2) and 4). It turns out for any linear transformation T, T ) = T ELFY - Use conjugate symmetry!). Thus Thus 1) and 2) are equivalent as well as 3) and 4). So we can just prove 1). It s pretty straightforward: y range T ) iff T x, y) = x, T y) = 0 for all x V. Also, x, T y) = 0 for all x V iff T y = 0, which means y Ker T This completes the proof Suggested Homework: 5.2. Hint: If P W is the orthogonal projection onto a subspace W of an inner product space V then I V P W is the orthogonal projection onto W.)

29 6. Isometries and Unitary Operators Let V and W be inner product spaces over F. T LV, W ) is an isometry if for all x V, T x = x. Proposition: T LV, W ) is an isometry iff for all x, y V, T x, T y) = x, y). Proof : Let s start with the reverse direction. Suppose for all x, y V, T x, T y) = x, y). Then for all x V, T x = T x, T x) = x, x) = x. So T is an isometry. Now assume T is an isometry. We split this direction into two cases, F = R and F = C: Assume F = R. Let x, y V. Then by the polarization identity, T x, T y) = 1 4 T x + T y 2 T x T y 2) = 1 4 T x + y) 2 T x y) 2) = = 1 4 x + y 2 x y 2) = x, y). Assume F = C. Let x, y V. Then by the polarization identity, T x, T y) = 1 4 T x + T y 2 T x T y 2 + i T x + it y 2 i T x it y 2) = = 1 4 T x + y) 2 T x y) 2 + i T x + iy) 2 i T x iy) 2) = = 1 4 x + y 2 x y 2 + i x + iy 2 i x iy 2) = x, y) So an isometry preserves the inner product.

30 Proposition: T LV, W ) is an isometry iff T T = I V. Proof : Let T LV, W ) be an isometry. Let y V. Then for all x V, x, T T y) = T x, T y) = x, y). Hence, T T y = y for all y V. Thus T T = I V. Now assume T T = I V. Let x, y V. Then T x, T y) = x, T T y) = x, I V y) = x, y). Thus T is an isometry The proposition above guarantees an isometry is left-invertible, but not necessarily invertible. An isometry T LV, W ) is a unitary operator if it is invertible. A matrix A M mn is an isometry if A A = I n. Also, a square matrix U M nn is unitary if U U = I n. ELFY) The columns of a matrix A M nn form an orthonormal basis iff A is unitary. If F = R then a unitary matrix is also called orthogonal. Basic properties of unitary matrices: Let U be a unitary matrix. 1) U 1 = U is also unitary. 2) If v 1, v 2,..., v n is an orthonormal basis of F n then Uv 1, Uv 2,..., Uv n is an orthonormal basis of F n. Suppose V, W have the same dimension. Let v 1, v 2,..., v n be an orthonormal basis of V and w 1, w 2,..., w n be an orthonormal basis of W. Define T LV, W ) by T v i = w i for i = 1, 2,...n. Then T is a unitary operator: Let x V. There exists c 1, c 2,..., c n F such that x = c 1 v 1 + c 2 v c n v n and x 2 = c c c n 2. Then n ) 2 T x 2 = T n 2 n 2 c i v i = c i T v i = c i w i = n c i 2 = x 2.

31 Proposition: Let U M nn be a unitary matrix. 1) det U = 1. For an orthogonal matrix U, det U = ±1.) 2) If λ is an eigenvalue of U then λ = 1. Proof : Let U M nn be a unitary matrix. Since det U = det U, it follows that 1 = det I n = detu U) = det U ) det U) = det Udet U = det U 2. Thus det U = 1, which proves 1). Let λ be an eigenvalue of U. Then there is a non-zero x F n such that Ux = λx. Then x = Ux = λx = λ x implies λ = 1, which proves 2) Two operators S, T LV ) or matrices S, T M nn ) are called unitarily equivalent if there exists and unitary operator U LV ) or matrix U M nn ) such that S = UT U. In particular, since U 1 = U, the operators/matrices would also be similar. However, not all similar operators/matrices are unitarily equivalent. Proposition: A M nn is unitarily equivalent to a diagonal matrix D iff A has an orthogonal basis of eigenvectors. Proof : Let A M nn. Assume A is unitarily equivalent to a diagonal matrix D. Then there is a unitary matrix U such that A = UDU. Since U U = I n, it is easy to check that the vectors Ue i for i = 1, 2,..., n are eigenvectors of A ELFY). Finally, since the vectors e 1, e 2,..., e n form an orthogonal orthonormal) basis, so do the eigenvectors Ue 1, Ue 2,..., Ue n since U is unitary. Now assume A has an orthogonal basis of eigenvectors v 1, v 2,..., v n. Let u i = 1 v i v i for i = 1, 2,..., n. Then B = {u 1, u 2,..., u n } is an orthonormal basis of eigenvectors of A. Let D = [A] BB and U be the matrix whose columns are u 1, u 2,..., u n. Clearly D is diagonal and U is unitary columns are an orthonormal basis). In particular, U = [I n ] SB, where S is the standard basis. Hence U = U 1 = [I n ] BS and A = [A] SS = [I n ] SB [A] BB [I n ] BS = UDU. So A is unitarily equivalent to D Suggested Homework: 6.1, 6.4, 6.6, and 6.8.

32 7. Rigid Motions in R n A rigid motion in an inner product space V is a transformation f : V V not necessarily linear) that preserves distance, that is, for all x, y V, fx) fy) = x y. Clearly every unitary operator T LV ) is a rigid motion. A nice example of a rigid motion that is not a linear operator is translation: Let y V be non-zero. Let f y : V V be given by f y x) = x + y. This operator is not linear since f0) 0. Proposition: Let f be a rigid motion in a real inner product space V. Let T : V V be given by T x) = fx) f0). Then for all x, y V, 1) T x) = x. 2) T x) T y) = x y. 3) T x), T y)) = x, y). Proof : Let f be a rigid motion in a real inner product space V. Let T : V V be given by T x) = fx) f0). Let x, y V. T x) = fx) f0) = x 0 = x. T x) T y) = fx) f0)) fy) f0)) = fx) fy) = x y. Since V is a real inner product space, it follows that x y 2 = x 2 2x, y) + y 2 and T x) T y) 2 = T x) 2 2T x), T y)) + T y) 2. By using 1) and 2) in the equations above, T x), T y)) = x, y)

33 The following proposition states that every rigid motion in a real inner product space is a composition of a translation and an orthogonal transformation. Proposition: Let f be a rigid motion in a real inner product space V. Let T : V V be given by T x) = fx) f0). T is an orthogonal transformation unitary operator in a real inner product space). Proof : Let x, y V and c R. Consider that T x + cy) T x) + ct y)) 2 = T x + cy) T x)) ct y)) 2 = = T x + cy) T x) 2 2cT x + cy) T x), T y)) + c 2 T y) 2 = = x + cy x 2 2cT x + cy), T y)) + 2cT x), T y)) + c 2 y 2 = = c 2 y 2 2cx + cy, y) + 2cx, y) + c 2 y 2 = = 2c 2 y 2 2cx, y) 2ccy, y) + 2cx, y) = 2c 2 y 2 2c 2 y, y) = 0. So T x + cy) T x) + ct y)) = 0, which means T x + cy) = T x) + ct y). Therefore T is linear, and thus by the previous proposition, T is an isometry. However, since T is an isometry from V to itself, it follows that T is invertible and thus unitary, which in a real inner product space is called an orthogonal transformation Note: We will not cover section 8 of this chapter.

### MATH 20F: LINEAR ALGEBRA LECTURE B00 (T. KEMP)

MATH 20F: LINEAR ALGEBRA LECTURE B00 (T KEMP) Definition 01 If T (x) = Ax is a linear transformation from R n to R m then Nul (T ) = {x R n : T (x) = 0} = Nul (A) Ran (T ) = {Ax R m : x R n } = {b R m

### Ir O D = D = ( ) Section 2.6 Example 1. (Bottom of page 119) dim(v ) = dim(l(v, W )) = dim(v ) dim(f ) = dim(v )

Section 3.2 Theorem 3.6. Let A be an m n matrix of rank r. Then r m, r n, and, by means of a finite number of elementary row and column operations, A can be transformed into the matrix ( ) Ir O D = 1 O

### Linear Algebra Massoud Malek

CSUEB Linear Algebra Massoud Malek Inner Product and Normed Space In all that follows, the n n identity matrix is denoted by I n, the n n zero matrix by Z n, and the zero vector by θ n An inner product

### Inner products. Theorem (basic properties): Given vectors u, v, w in an inner product space V, and a scalar k, the following properties hold:

Inner products Definition: An inner product on a real vector space V is an operation (function) that assigns to each pair of vectors ( u, v) in V a scalar u, v satisfying the following axioms: 1. u, v

### IMPORTANT DEFINITIONS AND THEOREMS REFERENCE SHEET

IMPORTANT DEFINITIONS AND THEOREMS REFERENCE SHEET This is a (not quite comprehensive) list of definitions and theorems given in Math 1553. Pay particular attention to the ones in red. Study Tip For each

### IMPORTANT DEFINITIONS AND THEOREMS REFERENCE SHEET

IMPORTANT DEFINITIONS AND THEOREMS REFERENCE SHEET This is a (not quite comprehensive) list of definitions and theorems given in Math 1553. Pay particular attention to the ones in red. Study Tip For each

### Math 18, Linear Algebra, Lecture C00, Spring 2017 Review and Practice Problems for Final Exam

Math 8, Linear Algebra, Lecture C, Spring 7 Review and Practice Problems for Final Exam. The augmentedmatrix of a linear system has been transformed by row operations into 5 4 8. Determine if the system

### Math Linear Algebra Final Exam Review Sheet

Math 15-1 Linear Algebra Final Exam Review Sheet Vector Operations Vector addition is a component-wise operation. Two vectors v and w may be added together as long as they contain the same number n of

### Eigenvectors and Hermitian Operators

7 71 Eigenvalues and Eigenvectors Basic Definitions Let L be a linear operator on some given vector space V A scalar λ and a nonzero vector v are referred to, respectively, as an eigenvalue and corresponding

### Math Camp Lecture 4: Linear Algebra. Xiao Yu Wang. Aug 2010 MIT. Xiao Yu Wang (MIT) Math Camp /10 1 / 88

Math Camp 2010 Lecture 4: Linear Algebra Xiao Yu Wang MIT Aug 2010 Xiao Yu Wang (MIT) Math Camp 2010 08/10 1 / 88 Linear Algebra Game Plan Vector Spaces Linear Transformations and Matrices Determinant

### 4.3 - Linear Combinations and Independence of Vectors

- Linear Combinations and Independence of Vectors De nitions, Theorems, and Examples De nition 1 A vector v in a vector space V is called a linear combination of the vectors u 1, u,,u k in V if v can be

### 2. Review of Linear Algebra

2. Review of Linear Algebra ECE 83, Spring 217 In this course we will represent signals as vectors and operators (e.g., filters, transforms, etc) as matrices. This lecture reviews basic concepts from linear

### The Spectral Theorem for normal linear maps

MAT067 University of California, Davis Winter 2007 The Spectral Theorem for normal linear maps Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (March 14, 2007) In this section we come back to the question

### Linear Algebra Final Exam Study Guide Solutions Fall 2012

. Let A = Given that v = 7 7 67 5 75 78 Linear Algebra Final Exam Study Guide Solutions Fall 5 explain why it is not possible to diagonalize A. is an eigenvector for A and λ = is an eigenvalue for A diagonalize

### Extra Problems for Math 2050 Linear Algebra I

Extra Problems for Math 5 Linear Algebra I Find the vector AB and illustrate with a picture if A = (,) and B = (,4) Find B, given A = (,4) and [ AB = A = (,4) and [ AB = 8 If possible, express x = 7 as

### Honors Linear Algebra, Spring Homework 8 solutions by Yifei Chen

.. Honors Linear Algebra, Spring 7. Homework 8 solutions by Yifei Chen 8... { { W {v R 4 v α v β } v x, x, x, x 4 x x + x 4 x + x x + x 4 } Solve the linear system above, we get a basis of W is {v,,,,

### 1. Select the unique answer (choice) for each problem. Write only the answer.

MATH 5 Practice Problem Set Spring 7. Select the unique answer (choice) for each problem. Write only the answer. () Determine all the values of a for which the system has infinitely many solutions: x +

### There are two things that are particularly nice about the first basis

Orthogonality and the Gram-Schmidt Process In Chapter 4, we spent a great deal of time studying the problem of finding a basis for a vector space We know that a basis for a vector space can potentially

### Stat 159/259: Linear Algebra Notes

Stat 159/259: Linear Algebra Notes Jarrod Millman November 16, 2015 Abstract These notes assume you ve taken a semester of undergraduate linear algebra. In particular, I assume you are familiar with the

### Mathematical Methods wk 2: Linear Operators

John Magorrian, magog@thphysoxacuk These are work-in-progress notes for the second-year course on mathematical methods The most up-to-date version is available from http://www-thphysphysicsoxacuk/people/johnmagorrian/mm

### Review of some mathematical tools

MATHEMATICAL FOUNDATIONS OF SIGNAL PROCESSING Fall 2016 Benjamín Béjar Haro, Mihailo Kolundžija, Reza Parhizkar, Adam Scholefield Teaching assistants: Golnoosh Elhami, Hanjie Pan Review of some mathematical

### 18.06 Problem Set 8 - Solutions Due Wednesday, 14 November 2007 at 4 pm in

806 Problem Set 8 - Solutions Due Wednesday, 4 November 2007 at 4 pm in 2-06 08 03 Problem : 205+5+5+5 Consider the matrix A 02 07 a Check that A is a positive Markov matrix, and find its steady state

### v = v 1 2 +v 2 2. Two successive applications of this idea give the length of the vector v R 3 :

Length, Angle and the Inner Product The length (or norm) of a vector v R 2 (viewed as connecting the origin to a point (v 1,v 2 )) is easily determined by the Pythagorean Theorem and is denoted v : v =

### Linear Algebra. Workbook

Linear Algebra Workbook Paul Yiu Department of Mathematics Florida Atlantic University Last Update: November 21 Student: Fall 2011 Checklist Name: A B C D E F F G H I J 1 2 3 4 5 6 7 8 9 10 xxx xxx xxx

### (a) If A is a 3 by 4 matrix, what does this tell us about its nullspace? Solution: dim N(A) 1, since rank(a) 3. Ax =

. (5 points) (a) If A is a 3 by 4 matrix, what does this tell us about its nullspace? dim N(A), since rank(a) 3. (b) If we also know that Ax = has no solution, what do we know about the rank of A? C(A)

### HOMEWORK PROBLEMS FROM STRANG S LINEAR ALGEBRA AND ITS APPLICATIONS (4TH EDITION)

HOMEWORK PROBLEMS FROM STRANG S LINEAR ALGEBRA AND ITS APPLICATIONS (4TH EDITION) PROFESSOR STEVEN MILLER: BROWN UNIVERSITY: SPRING 2007 1. CHAPTER 1: MATRICES AND GAUSSIAN ELIMINATION Page 9, # 3: Describe

### Elementary Linear Algebra Review for Exam 2 Exam is Monday, November 16th.

Elementary Linear Algebra Review for Exam Exam is Monday, November 6th. The exam will cover sections:.4,..4, 5. 5., 7., the class notes on Markov Models. You must be able to do each of the following. Section.4

### Vector Space Basics. 1 Abstract Vector Spaces. 1. (commutativity of vector addition) u + v = v + u. 2. (associativity of vector addition)

Vector Space Basics (Remark: these notes are highly formal and may be a useful reference to some students however I am also posting Ray Heitmann's notes to Canvas for students interested in a direct computational

### 5.) For each of the given sets of vectors, determine whether or not the set spans R 3. Give reasons for your answers.

Linear Algebra - Test File - Spring Test # For problems - consider the following system of equations. x + y - z = x + y + 4z = x + y + 6z =.) Solve the system without using your calculator..) Find the

### BASIC ALGORITHMS IN LINEAR ALGEBRA. Matrices and Applications of Gaussian Elimination. A 2 x. A T m x. A 1 x A T 1. A m x

BASIC ALGORITHMS IN LINEAR ALGEBRA STEVEN DALE CUTKOSKY Matrices and Applications of Gaussian Elimination Systems of Equations Suppose that A is an n n matrix with coefficents in a field F, and x = (x,,

### CHAPTER VIII HILBERT SPACES

CHAPTER VIII HILBERT SPACES DEFINITION Let X and Y be two complex vector spaces. A map T : X Y is called a conjugate-linear transformation if it is a reallinear transformation from X into Y, and if T (λx)

### Lecture 3: Review of Linear Algebra

ECE 83 Fall 2 Statistical Signal Processing instructor: R Nowak, scribe: R Nowak Lecture 3: Review of Linear Algebra Very often in this course we will represent signals as vectors and operators (eg, filters,

### NONCOMMUTATIVE POLYNOMIAL EQUATIONS. Edward S. Letzter. Introduction

NONCOMMUTATIVE POLYNOMIAL EQUATIONS Edward S Letzter Introduction My aim in these notes is twofold: First, to briefly review some linear algebra Second, to provide you with some new tools and techniques

### Linear Algebra Primer

Linear Algebra Primer David Doria daviddoria@gmail.com Wednesday 3 rd December, 2008 Contents Why is it called Linear Algebra? 4 2 What is a Matrix? 4 2. Input and Output.....................................

### 1 Invariant subspaces

MATH 2040 Linear Algebra II Lecture Notes by Martin Li Lecture 8 Eigenvalues, eigenvectors and invariant subspaces 1 In previous lectures we have studied linear maps T : V W from a vector space V to another

### 22m:033 Notes: 7.1 Diagonalization of Symmetric Matrices

m:33 Notes: 7. Diagonalization of Symmetric Matrices Dennis Roseman University of Iowa Iowa City, IA http://www.math.uiowa.edu/ roseman May 3, Symmetric matrices Definition. A symmetric matrix is a matrix

### LECTURE VI: SELF-ADJOINT AND UNITARY OPERATORS MAT FALL 2006 PRINCETON UNIVERSITY

LECTURE VI: SELF-ADJOINT AND UNITARY OPERATORS MAT 204 - FALL 2006 PRINCETON UNIVERSITY ALFONSO SORRENTINO 1 Adjoint of a linear operator Note: In these notes, V will denote a n-dimensional euclidean vector

### I. Multiple Choice Questions (Answer any eight)

Name of the student : Roll No : CS65: Linear Algebra and Random Processes Exam - Course Instructor : Prashanth L.A. Date : Sep-24, 27 Duration : 5 minutes INSTRUCTIONS: The test will be evaluated ONLY

### Math 25a Practice Final #1 Solutions

Math 25a Practice Final #1 Solutions Problem 1. Suppose U and W are subspaces of V such that V = U W. Suppose also that u 1,..., u m is a basis of U and w 1,..., w n is a basis of W. Prove that is a basis

### MATH 369 Linear Algebra

Assignment # Problem # A father and his two sons are together 00 years old. The father is twice as old as his older son and 30 years older than his younger son. How old is each person? Problem # 2 Determine

### which arises when we compute the orthogonal projection of a vector y in a subspace with an orthogonal basis. Hence assume that P y = A ij = x j, x i

MODULE 6 Topics: Gram-Schmidt orthogonalization process We begin by observing that if the vectors {x j } N are mutually orthogonal in an inner product space V then they are necessarily linearly independent.

### Homework 2. Solutions T =

Homework. s Let {e x, e y, e z } be an orthonormal basis in E. Consider the following ordered triples: a) {e x, e x + e y, 5e z }, b) {e y, e x, 5e z }, c) {e y, e x, e z }, d) {e y, e x, 5e z }, e) {

### ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA

ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND

### OPERATOR THEORY ON HILBERT SPACE. Class notes. John Petrovic

OPERATOR THEORY ON HILBERT SPACE Class notes John Petrovic Contents Chapter 1. Hilbert space 1 1.1. Definition and Properties 1 1.2. Orthogonality 3 1.3. Subspaces 7 1.4. Weak topology 9 Chapter 2. Operators

### Lecture 23: 6.1 Inner Products

Lecture 23: 6.1 Inner Products Wei-Ta Chu 2008/12/17 Definition An inner product on a real vector space V is a function that associates a real number u, vwith each pair of vectors u and v in V in such

### Supplementary Notes on Linear Algebra

Supplementary Notes on Linear Algebra Mariusz Wodzicki May 3, 2015 1 Vector spaces 1.1 Coordinatization of a vector space 1.1.1 Given a basis B = {b 1,..., b n } in a vector space V, any vector v V can

### Linear Algebra 2 Final Exam, December 7, 2015 SOLUTIONS. a + 2b = x a + 3b = y. This solves to a = 3x 2y, b = y x. Thus

Linear Algebra 2 Final Exam, December 7, 2015 SOLUTIONS 1. (5.5 points) Let T : R 2 R 4 be a linear mapping satisfying T (1, 1) = ( 1, 0, 2, 3), T (2, 3) = (2, 3, 0, 0). Determine T (x, y) for (x, y) R

### Chapter Two Elements of Linear Algebra

Chapter Two Elements of Linear Algebra Previously, in chapter one, we have considered single first order differential equations involving a single unknown function. In the next chapter we will begin to

### Math 224, Fall 2007 Exam 3 Thursday, December 6, 2007

Math 224, Fall 2007 Exam 3 Thursday, December 6, 2007 You have 1 hour and 20 minutes. No notes, books, or other references. You are permitted to use Maple during this exam, but you must start with a blank

### The Gram Schmidt Process

u 2 u The Gram Schmidt Process Now we will present a procedure, based on orthogonal projection, that converts any linearly independent set of vectors into an orthogonal set. Let us begin with the simple

### The Gram Schmidt Process

The Gram Schmidt Process Now we will present a procedure, based on orthogonal projection, that converts any linearly independent set of vectors into an orthogonal set. Let us begin with the simple case

### (VII.E) The Singular Value Decomposition (SVD)

(VII.E) The Singular Value Decomposition (SVD) In this section we describe a generalization of the Spectral Theorem to non-normal operators, and even to transformations between different vector spaces.

### Math 396. An application of Gram-Schmidt to prove connectedness

Math 396. An application of Gram-Schmidt to prove connectedness 1. Motivation and background Let V be an n-dimensional vector space over R, and define GL(V ) to be the set of invertible linear maps V V

### Some notes on Coxeter groups

Some notes on Coxeter groups Brooks Roberts November 28, 2017 CONTENTS 1 Contents 1 Sources 2 2 Reflections 3 3 The orthogonal group 7 4 Finite subgroups in two dimensions 9 5 Finite subgroups in three

### Vector Geometry. Chapter 5

Chapter 5 Vector Geometry In this chapter we will look more closely at certain geometric aspects of vectors in R n. We will first develop an intuitive understanding of some basic concepts by looking at

### 4.1 Distance and Length

Chapter Vector Geometry In this chapter we will look more closely at certain geometric aspects of vectors in R n. We will first develop an intuitive understanding of some basic concepts by looking at vectors

### CAAM 336 DIFFERENTIAL EQUATIONS IN SCI AND ENG Examination 1

CAAM 6 DIFFERENTIAL EQUATIONS IN SCI AND ENG Examination Instructions: Time limit: uninterrupted hours There are four questions worth a total of 5 points Please do not look at the questions until you begin

### Real symmetric matrices/1. 1 Eigenvalues and eigenvectors

Real symmetric matrices 1 Eigenvalues and eigenvectors We use the convention that vectors are row vectors and matrices act on the right. Let A be a square matrix with entries in a field F; suppose that

### Math 443 Differential Geometry Spring Handout 3: Bilinear and Quadratic Forms This handout should be read just before Chapter 4 of the textbook.

Math 443 Differential Geometry Spring 2013 Handout 3: Bilinear and Quadratic Forms This handout should be read just before Chapter 4 of the textbook. Endomorphisms of a Vector Space This handout discusses

### 18.06 Problem Set 10 - Solutions Due Thursday, 29 November 2007 at 4 pm in

86 Problem Set - Solutions Due Thursday, 29 November 27 at 4 pm in 2-6 Problem : (5=5+5+5) Take any matrix A of the form A = B H CB, where B has full column rank and C is Hermitian and positive-definite

### Ph.D. Katarína Bellová Page 1 Mathematics 2 (10-PHY-BIPMA2) EXAM - Solutions, 20 July 2017, 10:00 12:00 All answers to be justified.

PhD Katarína Bellová Page 1 Mathematics 2 (10-PHY-BIPMA2 EXAM - Solutions, 20 July 2017, 10:00 12:00 All answers to be justified Problem 1 [ points]: For which parameters λ R does the following system

### Problem Set 1. Homeworks will graded based on content and clarity. Please show your work clearly for full credit.

CSE 151: Introduction to Machine Learning Winter 2017 Problem Set 1 Instructor: Kamalika Chaudhuri Due on: Jan 28 Instructions This is a 40 point homework Homeworks will graded based on content and clarity

### Vector Calculus. Lecture Notes

Vector Calculus Lecture Notes Adolfo J. Rumbos c Draft date November 23, 211 2 Contents 1 Motivation for the course 5 2 Euclidean Space 7 2.1 Definition of n Dimensional Euclidean Space........... 7 2.2

### Diagonalization by a unitary similarity transformation

Physics 116A Winter 2011 Diagonalization by a unitary similarity transformation In these notes, we will always assume that the vector space V is a complex n-dimensional space 1 Introduction A semi-simple

### Linear algebra 2. Yoav Zemel. March 1, 2012

Linear algebra 2 Yoav Zemel March 1, 2012 These notes were written by Yoav Zemel. The lecturer, Shmuel Berger, should not be held responsible for any mistake. Any comments are welcome at zamsh7@gmail.com.

### Math 225 Linear Algebra II Lecture Notes. John C. Bowman University of Alberta Edmonton, Canada

Math 225 Linear Algebra II Lecture Notes John C Bowman University of Alberta Edmonton, Canada March 23, 2017 c 2010 John C Bowman ALL RIGHTS RESERVED Reproduction of these lecture notes in any form, in

### MATH 304 Linear Algebra Lecture 23: Diagonalization. Review for Test 2.

MATH 304 Linear Algebra Lecture 23: Diagonalization. Review for Test 2. Diagonalization Let L be a linear operator on a finite-dimensional vector space V. Then the following conditions are equivalent:

### Math 113 Homework 5. Bowei Liu, Chao Li. Fall 2013

Math 113 Homework 5 Bowei Liu, Chao Li Fall 2013 This homework is due Thursday November 7th at the start of class. Remember to write clearly, and justify your solutions. Please make sure to put your name

### MA 265 FINAL EXAM Fall 2012

MA 265 FINAL EXAM Fall 22 NAME: INSTRUCTOR S NAME:. There are a total of 25 problems. You should show work on the exam sheet, and pencil in the correct answer on the scantron. 2. No books, notes, or calculators

### Lecture 11: Finish Gaussian elimination and applications; intro to eigenvalues and eigenvectors (1)

Lecture 11: Finish Gaussian elimination and applications; intro to eigenvalues and eigenvectors (1) Travis Schedler Tue, Oct 18, 2011 (version: Tue, Oct 18, 6:00 PM) Goals (2) Solving systems of equations

### Math 314/ Exam 2 Blue Exam Solutions December 4, 2008 Instructor: Dr. S. Cooper. Name:

Math 34/84 - Exam Blue Exam Solutions December 4, 8 Instructor: Dr. S. Cooper Name: Read each question carefully. Be sure to show all of your work and not just your final conclusion. You may not use your

### Preliminary Linear Algebra 1. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 100

Preliminary Linear Algebra 1 Copyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 1 / 100 Notation for all there exists such that therefore because end of proof (QED) Copyright c 2012

### ALGEBRA 8: Linear algebra: characteristic polynomial

ALGEBRA 8: Linear algebra: characteristic polynomial Characteristic polynomial Definition 8.1. Consider a linear operator A End V over a vector space V. Consider a vector v V such that A(v) = λv. This

### Optimization Theory. A Concise Introduction. Jiongmin Yong

October 11, 017 16:5 ws-book9x6 Book Title Optimization Theory 017-08-Lecture Notes page 1 1 Optimization Theory A Concise Introduction Jiongmin Yong Optimization Theory 017-08-Lecture Notes page Optimization

### Image Registration Lecture 2: Vectors and Matrices

Image Registration Lecture 2: Vectors and Matrices Prof. Charlene Tsai Lecture Overview Vectors Matrices Basics Orthogonal matrices Singular Value Decomposition (SVD) 2 1 Preliminary Comments Some of this

### Hilbert spaces. 1. Cauchy-Schwarz-Bunyakowsky inequality

(October 29, 2016) Hilbert spaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/fun/notes 2016-17/03 hsp.pdf] Hilbert spaces are

### The Gram-Schmidt Process 1

The Gram-Schmidt Process In this section all vector spaces will be subspaces of some R m. Definition.. Let S = {v...v n } R m. The set S is said to be orthogonal if v v j = whenever i j. If in addition

### Review and problem list for Applied Math I

Review and problem list for Applied Math I (This is a first version of a serious review sheet; it may contain errors and it certainly omits a number of topic which were covered in the course. Let me know

### Solutions to Final Practice Problems Written by Victoria Kala Last updated 12/5/2015

Solutions to Final Practice Problems Written by Victoria Kala vtkala@math.ucsb.edu Last updated /5/05 Answers This page contains answers only. See the following pages for detailed solutions. (. (a x. See

### Problems in Linear Algebra and Representation Theory

Problems in Linear Algebra and Representation Theory (Most of these were provided by Victor Ginzburg) The problems appearing below have varying level of difficulty. They are not listed in any specific

### 5. Orthogonal matrices

L Vandenberghe EE133A (Spring 2017) 5 Orthogonal matrices matrices with orthonormal columns orthogonal matrices tall matrices with orthonormal columns complex matrices with orthonormal columns 5-1 Orthonormal

### Topics in linear algebra

Chapter 6 Topics in linear algebra 6.1 Change of basis I want to remind you of one of the basic ideas in linear algebra: change of basis. Let F be a field, V and W be finite dimensional vector spaces over

### Math 702 Problem Set 10 Solutions

Math 70 Problem Set 0 Solutions Unless otherwise stated, X is a complex Hilbert space with the inner product h; i the induced norm k k. The null-space range of a linear map T W X! X are denoted by N.T

### Solutions to Math 51 First Exam April 21, 2011

Solutions to Math 5 First Exam April,. ( points) (a) Give the precise definition of a (linear) subspace V of R n. (4 points) A linear subspace V of R n is a subset V R n which satisfies V. If x, y V then

### Linear algebra II Homework #1 solutions A = This means that every eigenvector with eigenvalue λ = 1 must have the form

Linear algebra II Homework # solutions. Find the eigenvalues and the eigenvectors of the matrix 4 6 A =. 5 Since tra = 9 and deta = = 8, the characteristic polynomial is f(λ) = λ (tra)λ+deta = λ 9λ+8 =

### 33A Linear Algebra and Applications: Practice Final Exam - Solutions

33A Linear Algebra and Applications: Practice Final Eam - Solutions Question Consider a plane V in R 3 with a basis given by v = and v =. Suppose, y are both in V. (a) [3 points] If [ ] B =, find. (b)

### THE EULER CHARACTERISTIC OF A LIE GROUP

THE EULER CHARACTERISTIC OF A LIE GROUP JAY TAYLOR 1 Examples of Lie Groups The following is adapted from [2] We begin with the basic definition and some core examples Definition A Lie group is a smooth

### 2.2. OPERATOR ALGEBRA 19. If S is a subset of E, then the set

2.2. OPERATOR ALGEBRA 19 2.2 Operator Algebra 2.2.1 Algebra of Operators on a Vector Space A linear operator on a vector space E is a mapping L : E E satisfying the condition u, v E, a R, L(u + v) = L(u)

### Solutions for Math 225 Assignment #5 1

Solutions for Math 225 Assignment #5 1 (1) Find a polynomial f(x) of degree at most 3 satisfying that f(0) = 2, f( 1) = 1, f(1) = 3 and f(3) = 1. Solution. By Lagrange Interpolation, ( ) (x + 1)(x 1)(x

### Boolean Inner-Product Spaces and Boolean Matrices

Boolean Inner-Product Spaces and Boolean Matrices Stan Gudder Department of Mathematics, University of Denver, Denver CO 80208 Frédéric Latrémolière Department of Mathematics, University of Denver, Denver

### Department of Biostatistics Department of Stat. and OR. Refresher course, Summer Linear Algebra. Instructor: Meilei Jiang (UNC at Chapel Hill)

Department of Biostatistics Department of Stat. and OR Refresher course, Summer 216 Linear Algebra Original Author: Oleg Mayba (UC Berkeley, 26) Modified By: Eric Lock (UNC, 21 & 211) Gen Li (UNC, 212)

### Math 110 Linear Algebra Midterm 2 Review October 28, 2017

Math 11 Linear Algebra Midterm Review October 8, 17 Material Material covered on the midterm includes: All lectures from Thursday, Sept. 1st to Tuesday, Oct. 4th Homeworks 9 to 17 Quizzes 5 to 9 Sections

### Linear Algebra Problems

Math 504 505 Linear Algebra Problems Jerry L. Kazdan Topics 1 Basics 2 Linear Equations 3 Linear Maps 4 Rank One Matrices 5 Algebra of Matrices 6 Eigenvalues and Eigenvectors 7 Inner Products and Quadratic

### Chapter 3. Matrices. 3.1 Matrices

40 Chapter 3 Matrices 3.1 Matrices Definition 3.1 Matrix) A matrix A is a rectangular array of m n real numbers {a ij } written as a 11 a 12 a 1n a 21 a 22 a 2n A =.... a m1 a m2 a mn The array has m rows

### Inner Product Spaces 5.2 Inner product spaces

Inner Product Spaces 5.2 Inner product spaces November 15 Goals Concept of length, distance, and angle in R 2 or R n is extended to abstract vector spaces V. Sucn a vector space will be called an Inner

### Algebra I Fall 2007

MIT OpenCourseWare http://ocw.mit.edu 18.701 Algebra I Fall 007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.701 007 Geometry of the Special Unitary

### Topic 2 Quiz 2. choice C implies B and B implies C. correct-choice C implies B, but B does not imply C

Topic 1 Quiz 1 text A reduced row-echelon form of a 3 by 4 matrix can have how many leading one s? choice must have 3 choice may have 1, 2, or 3 correct-choice may have 0, 1, 2, or 3 choice may have 0,

### Linear Algebra in Actuarial Science: Slides to the lecture

Linear Algebra in Actuarial Science: Slides to the lecture Fall Semester 2010/2011 Linear Algebra is a Tool-Box Linear Equation Systems Discretization of differential equations: solving linear equations