8.024 HW2 Solutions February 4, 200 Apostol.3 4,7,8 4. Show that for all x and y in real Euclidean space: (x, y) 0 x + y 2 x 2 + y 2 Proof. By definition of the norm and the linearity of the inner product, for any x, y the following holds: x+y 2 (x+y, x+y) (x, x+y)+(y, x+y) (x, x)+(x, y)+(y, x)+(y, y) x 2 + y 2 +(x, y)+(y, x). Because this is a real vector space, (x, y) (y, x) and this becomes x + y 2 x 2 + y 2 + 2(x, y) (0.) which holds for ANY x, y. So, if (x, y) 0 then 0. becomes x + y 2 x 2 + y 2. On the other hand, if x + y 2 x 2 + y 2, subtract this equation from 0.. This gives: which implies that (x, y) 0. 0 2(x, y) 7. Prove that if x and y are nonzero elements making an angle θ with each other, then x y 2 x 2 + y 2 2 x y cos θ. Proof. Recall (Definition p 8) that in real Euclidean space the angle θ between two nonzero x and y is defined to be the number in the interval 0 θ π satisfying cos θ Using again the definition of the norm and its linearity, we have (x, y) x y. (0.2) x y 2 (x y, x y) (x, x y) (y, x y) (x, x) (x, y) (y, x)+(y, y) x 2 + y 2 2(x, y). By equation 0.2, (x, y) x y cos θ so replacing (x, y) with this in the expression above gives the desired equality.
8. In the real linear space C(, e) define an inner product by the equation (f, g) e (log x)f(x)g(x)dx (a) If f(x) x, compute f. and so f 2 (f, f) f e x(log x)dx 4 ( + e2 ) 4 ( + e2 ) + e 2 2 (b) Find a linear polynomial g(x) a + bx that is orthogonal to the constant function f(x). Let g(x) a + bx be an arbitrary linear polynomial, then its inner product with f(x) is equal to (g, f) e (log x)(a + bx)dx a + b 4 ( + e2 ). So, for g to be orthogonal to f, it must satisfy a + b 4 ( + e2 ) 0, or a b 4 ( + e2 ). One possible such choice is b 4, a ( + e 2 ). Can plug this in an explicitly compute that its orthogonal. 2. Apostol 2.2 2.c, 3 2.c Determine the matrix for the projection T : V 5 V 3 where T (x, x 2, x 3, x 4, x 5 ) (x 2, x 3, x 4 ). Write e, e 2, e 3, e 4, e 5 for the standard basis elements in V 5, ie e (, 0, 0, 0, 0), etc. Write ẽ, ẽ 2, ẽ 3 for the standard basis elements in V 3, ie ẽ (, 0, 0), etc. Look at where each basis element e i maps to in V 3. Write t k, t 2k, t 3k for the components of T (e k ) relative to the ordered basis (ẽ, ẽ 2, ẽ 3 ), meaning that T (e k ) 3 t ik ẽ i T (e ) 0 0 ẽ + 0 ẽ 2 + 0 ẽ 3, so t t 2 t 3 0. T (e 2 ) (, 0, 0) e so t 2 and t 22 t 32 0. T (e 3 ) (0,, 0) e 2 so t 23 and t 3 t 33 0. T (e 4 ) (0, 0, ) e 3 so t 34 and t 4 t 24 0. T (e 5 ) 0 so t 5 t 25 t 35 0. So the matrix representation of T is i 0 0 0 0 0 0 0 0 0 0 0 0 3. A linear transformation T : V 2 V 2 maps the basis vectors i, j as follows: T (i) i + j 2
T (j) 2i j (a) Compute T (3i 4j) and T 2 (3i 4j) in terms of i and j. T (3i 4j) 3T (i) 4T (j) 3(i + j) 4(2i j) 5i + 7j T 2 (3i 4j) T (T (3i 4j)) T ( 5i + 7j) 5T (i) + 7T (j) 5(i + j) + 7(2i j) 9i 2j (b) Determine the matrix of T and T 2 : We know that T (i) i + j so the [ corresponding column 2 vector is and T (j) 2i j so the corresponding column vector is and combining these the transformation matrix for T is [ 2 Similarly, T 2 (i) T (T (i)) T (i + j) T (i) + T (j) (i + j) + (2i j) 3i T 2 (j) T (T (j)) T (2i j) 2T (i) T (j) 2(i + j) (2i j) 3j so the transformation matrix is [ 3 0 0 3 (c.) Solve part b. if the basis (i, j) is replaced by (e, e 2 ) where e i j and e 2 3i + j. We must rewrite T (i) and T (j) as a sum of the new basis vectors and take note of the coefficients: T (e ) T (i j) T (i) T (j) (i + j) (2i j) i + 2j 7 4 (i j) + 4 (3i + j) 7 4 e + 4 e 2 T (e 2 ) T (3i+j) 3T (i)+t (j) 3(i+j)+(2i j) 5i+2j 4 (i j)+ 7 4 (3i+j) 4 e + 7 4 e 2 so the transformation matrix is [ 7 Also: 4 4 7 4 4 T 2 (e ) T ( i + 2j) T (i) + 2T (j) (i + j) + 2(2i j) 3i 3j 3(i j) 3e T 2 (e 2 ) T (5i + 2j) 5T (i) + 2T (j) 5(i + j) + 2(2i j) 9i + 3j 3(3i + j) 3e 2 so the transformation matrix is [ 3 0 0 3 3
[ cos θ sin θ 3 Apostol 2.6 7 Let A sin θ cos θ A n. [ cos 2θ sin 2θ and verify that A 2 sin 2θ cos 2θ. Compute Compute A 2 explicitly by multiplying A A: A 2 cos θ sin θ cos θ sin θ sin θ cos θ sin θ cos θ cos 2θ sin 2θ sin 2θ cos 2θ [ cos 2 θ sin 2 θ 2 sin θ cos θ 2 sin θ cos θ sin 2 θ + cos 2 θ where the final equality follows from the double angle formula from trigonometry. cos nθ sin nθ Claim 0.. A n sin nθ cos nθ Proof. We will prove this by induction on n. We may take n as our base case, and see that the formula holds. Now assume that it holds for n, ie assume that A n cos(n )θ sin(n )θ sin(n )θ cos(n )θ and show that it holds for A n. We compute A n A A n as follows: A n cos θ sin θ cos(n )θ sin(n )θ sin θ cos θ sin(n )θ cos(n )θ [ cos θ cos(n )θ sin θ sin(n )θ cos θ sin(n )θ sin θ cos(n )θ sin θ cos(n )θ + cos θ sin(n )θ sin θ sin(n )θ + cos θ cos(n )θ Now use the angle sum identities from trigonometry (of which the double angle formula is a special case): sin(α + β) sin α cos β + cos α sin β Using these in the matrix above gives cos(α + β) cos α cos β sin α sin β. [ cos nθ sin nθ sin nθ cos nθ which is what we wanted to show. 4 Apostol 2.2 2.b,c,e,f For each of the following statements about n n matrices give a proof or exhibit a counter example. (b) If A and B are nonsingular, then A + B is nonsingular. False! For example, let A [ 0 0 then A + B is the zero matrix which is not invertible. [ 0, B 0 (c) If A and B are nonsingular, then AB is nonsingular. 4
True! Working with determinants, we have that A and B are nonsingular implies that det A 0 and det B 0. Since det AB det A det B 0, this implies that AB is also nonsingular. Alternatively, if we let A denote the inverse of A and B denote the inverse of B, then so B A is the inverse of AB. AB (B A ) A(BB )A AA I (e) If A 3 0 then A I is nonsingular. True! We prove this by giving an explicit inverse for A I. Let B (A 2 + A + I), then (A I)B (A 3 + A 2 + A A 2 A I) I (f) If the product of k matrices A...A k is nonsingular, then each matrix A i is nonsingular. True! Since A...A k is nonsingular, 0 det(a...a k ) det A... det A k. If any matrix A i was not invertible, then det A i 0 and so the product of all the determinants would equal 0, a contradiction. 5 Apostol 3.6 2.b, 3.a x y z 2.b Let A 3 0 2 and assume that det A. Show that the determinant of x y z B 3x + 3 3y 3z + 2 x + y + z + is. Write the rows of A as A, A 2, A 3 and the rows of B as B, B 2, B 3. Comparing the two matrices, we see that B A, B 2 3A + A 2 and B 3 A + A 3, so matrix B can be gotten from matrix A by adding a scalar multiple of the first row to each of the other two rows. This does not change the determinant (see the axiomatic definition of determinants as well as Example 4 p. 78), so det B det A. 3(b) Let Prove that A a b c a 2 b 2 c 2 det A (b a)(c a)(c b) We will use the Gauss-Jordan process as in Example 4 p. 78 to get an upper triangular matrix, which we know how to compute the determinant of by Example 3 p 77 (product of the diagonal elements). For any matrix M, write the rows of M as M, M 2, M 3. Step : Subtract aa from A 2. This does not affect the determinant. The resulting matrix is A 0 b a c a a 2 b 2 c 2 5
Step 2: Subtract a 2 A from A 3, which again does not affect the determinant. The result is A 0 b a c a 0 b 2 a 2 c 2 a 2 Finally, subtract (b + a)a 2 from A 3 to get the matrix: 0 b a c a 0 0 (c 2 a 2 ) (b + a)(c a) 0 b a c a 0 0 (c a)(c b) The determinant of this matrix is (b a)(c a)(c b). Since none of the operations that we performed affect the determinant, this shows that the determinant of A is also (b a)(c a)(c b). 6 H Let A be a square n n matrix. Compute det(cof(a)) in terms of det A. Claim 0.2. det(cof(a)) (det A) n Proof. By Theorem 3.2, we have that A det A (cofa)t and also that det A det A t. Combining these, we have det(cofa) det(cofa) t det(det A A ) (det A) n (det A) (det A) n 7 H2 Let A be a square matrix all of whose entries are integers. Show that if det A + or then the entries of A are also integers. For an arbitrary n find an n n matrix A all of whose entries are non-zero integers and det A. Again we use the fact here that A det A (cofa)t so if det A ±,, then A ±(cofa) t. Since each entry in the cofactor matrix is of the form cofa ij ( ) i+j det A ij, these will be integers if det A ij is. But this is just some sum of products of entries in A, which are all integers by assumption. So each cofa ij is an integer and so A has all integer entries. There are lots of examples of matrices whose entries allall nonzero integers... 0... and det A. For example, if you take the n n upper triangular matrix.......... 0 0... which has s on the diagonal and in every spot above the diagonal, and zeros below. Then the determinant of this matrix is. Now, we can add scalar multiples of rows to other rows to get another matrix with no zero entries but that still has determinant. For example, add the first... 2... 2 row to each of the other rows to get............. 2 6