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

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1 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 = (λ )(λ 8). The eigenvectors with eigenvalue λ = satisfy the system Av = v, namely 3 6 x (A I)v = = = = x+y = = x = y. 4 y This means that every eigenvector with eigenvalue λ = must have the form y v = = y, y. y Similarly, the eigenvectors with eigenvalue λ = 8 are solutions of Av = 8v, so 4 6 x (A 8I)v = = = = x 3y = = x = 3y/ 3 y and every eigenvector with eigenvalue λ = 8 must have the form 3y/ 3/ v = = y, y. y. Find the eigenvalues and the eigenvectors of the matrix 3 3 A = The eigenvalues of A are the roots of the characteristic polynomial f(λ) = det(a λi) = λ 3 +7λ 6λ+ = (λ ) (λ 3). The eigenvectors of A are nonzero vectors in the null spaces 3 N(A I) = Span,, N(A 3I) = Span. 3

2 3. The following matrix has eigenvalues λ =,,,4. Is it diagonalisable? Explain A = When it comes to the eigenvalue λ =, row reduction of A λi gives 3 3 A λi = , so there are pivots and linearly independent eigenvectors. When λ = 4, we similarly get 3 3 A λi = , 3 3 so there are 3 pivots and linearly independent eigenvector. This gives a total of 3 linearly independent eigenvectors, so A is not diagonalisable. One does not really need to find the eigenvectors in this case, but those are nonzero elements of the null spaces N(A I) = Span,, N(A 4I) = Span. 4. Suppose A is a square matrix which is both diagonalisable and invertible. Show that every eigenvector of A is an eigenvector of A and that A is diagonalisable. Suppose that v is an eigenvector of A with eigenvalue λ. Then λ is nonzero because λ = = Av = λv = = A Av = = v = and eigenvectors are nonzero. To see that v is also an eigenvector of A, we note that λv = Av = λa v = v = A v = λ v. This proves the first part. To prove the second, assume that A is an n n matrix. It must then have n linearly independent eigenvectors which form a matrix B such that B AB is diagonal. These vectors are also eigenvectors of A, so B A B is diagonal as well.

3 Linear algebra II Homework # solutions. Find a basis for both the null space and the column space of the matrix 5 A = The reduced row echelon form of A is easily found to be R = 3. Since the pivots appear in the first and the second columns, this implies that C(A) = Span{Ae,Ae } = Span 3 4,. 4 3 The null space of A is the same as the null space of R. It can be expressed in the form x 3 +x 4 N(A) = x 3 3x 4 x 3 : x 3,x 4 R = Span, 3. x 4. Show that the following matrix is diagonalisable. 7 3 A = The eigenvalues of A are the roots of the characteristic polynomial f(λ) = det(a λi) = λ 3 +λ 38λ+4 = (λ )(λ 4)(λ 5). Since the eigenvalues of A are distinct, we conclude that A is diagonalisable.

4 3. Find the eigenvalues and the generalised eigenvectors of the matrix A = The eigenvalues of A are the roots of the characteristic polynomial f(λ) = det(a λi) = λ 3 +6λ 9λ+4 = (4 λ)(λ ). When it comes to the eigenvalue λ = 4, one can easily check that N(A 4I) = Span, N(A 4I) = N(A 4I). This implies that N(A 4I) j = N(A 4I) for all j, so we have found all generalised eigenvectors with λ = 4. When it comes to the eigenvalue λ =, one similarly has N(A I) = Span, N(A I) = N(A I) 3 = Span,. In view of the general theory, we must thus have N(A I) j = N(A I) for all j. 4. Suppose that v,v,...,v k form a basis for the null space of a square matrix A and that C = B AB for some invertible matrix B. Find a basis for the null space of C. We note that a vector v lies in the null space of C = B AB if and only if B ABv = ABv = Bv N(A) Bv = k c i v i i= for some scalar coefficients c i. In particular, the vectors B v i span the null space of C. To show that these vectors are also linearly independent, we note that k c i B v i = i= k c i v i = c i = for all i. i=

5 Linear algebra II Homework #3 solutions. The following matrix has λ = 4 as its only eigenvalue. What is its Jordan form? 5 A = 7. 4 In this case, the null space of A 4I is one-dimensional, as row reduction gives 3 5 A 4I = 3. Similarly, the null space of (A 4I) is two-dimensional because (A 4I) =, while (A 4I) 3 is the zero matrix, so its null space is three-dimensional. The diagram of Jordan chains is then and there is a single 3 3 Jordan block, namely λ 4 J = λ = 4. λ 4. The following matrix has λ = 4 as its only eigenvalue. What is its Jordan form? 3 3 A = 6. 5 In this case, the null space of A 4I is two-dimensional, as row reduction gives A 4I =. On the other hand, (A 4I) is the zero matrix, so its null space is three-dimensional. Thus, the diagram of Jordan chains is and there is a Jordan chain of length as well as a Jordan chain of length. These Jordan chains give a block and an block, so 4 J = 4. 4

6 3. Find a Jordan chain of length for the matrix 5 A =. The eigenvalues of the given matrix are the roots of the characteristic polynomial f(λ) = λ (tra)λ+deta = λ 6λ+9 = (λ 3), so λ = 3 is the only eigenvalue. Using row reduction, we now get A 3I =, so the null space of A 3I is one-dimensional. On the other hand, (A 3I) is the zero matrix, so its null space is two-dimensional. To find a Jordan chain of length, we pick a vector v that lies in the latter null space, but not in the former. We can always take v = e = = v = (A 3I)v =, but there are obviously infinitely many choices. Another possible choice would be v = e = = v = (A 3I)v =. 4. Let A be a 3 3 matrix that has v,v,v 3 as a Jordan chain of length 3 and let B be the matrix whose columns are v 3,v,v (in the order listed). Compute B AB. Suppose the vectors v,v,v 3 form a Jordan chain with eigenvalue λ, in which case (A λi)v = v, (A λi)v = v 3, (A λi)v 3 =. To find the columns of B AB, we need to find the coefficients that one needs in order to express the vectors Av 3,Av,Av in terms of the given basis. By above, we have Av 3 = λv 3, Av = v 3 +λv, Av = v +λv. Reading the coefficients of the vectors v 3,v,v in the given order, we conclude that λ B AB = λ. λ

7 Linear algebra II Homework #4 solutions. Find the Jordan form and a Jordan basis for the matrix 3 A = The characteristic polynomial of the given matrix is f(λ) = det(a λi) = λ 3 +4λ 5λ+ = ( λ)(λ ), so its eigenvalues are λ =,,. The corresponding null spaces are easily found to be N(A I) = Span,, N(A I) = Span 3. These contain 3 linearly independent eigenvectors, so A is diagonalisable and B = 3 = J = B AB =.. Find the Jordan form and a Jordan basis for the matrix 4 A = 4. The characteristic polynomial of the given matrix is f(λ) = det(a λi) = λ 3 +8λ λ+6 = (4 λ)(λ ), so its eigenvalues are λ = 4,,. The corresponding null spaces are easily found to be N(A 4I) = Span, N(A I) = Span. This implies that A is not diagonalisable and that its Jordan form is 4 J = B AB =.

8 To find a Jordan basis, we need to find vectors v,v,v 3 such that v is an eigenvector with eigenvalue λ = 4 and v,v 3 is a Jordan chain with eigenvalue λ =. In our case, we have N(A I) = Span,, so it easily follows that a Jordan basis is provided by the vectors v =, v =, v 3 = (A I)v =. 3. Suppose that A is a matrix with A = A. Show that A is diagonalisable. Suppose that A is not diagonalisable. Then its eigenvalues are not distinct, so there is a double eigenvalue λ and the Jordan form is λ λ λ λ J = = J = = λ λ λ λ λ. Write J = B AB for some invertible matrix B. Since A = A, we must also have λ J = B AB B AB = B A B = B AB = J =. λ Comparing the last two equations now gives λ = λ and λ =, a contradiction. 4. A 5 5 matrix A has characteristic polynomial f(λ) = λ 4 ( λ) and its column space is two-dimensional. Find the dimension of the column space of A. The eigenvalue λ = has multiplicity 4 and the number of Jordan chains is dimn(a) = 5 dimc(a) = 3. In particular, the Jordan chain diagram is and we have dimn(a ) = 4, so dimc(a ) = 5 dimn(a ) =.

9 Linear algebra II Homework #5 solutions. Let x = and y =. Suppose the sequences x n,y n are such that x n = 4x n y n, y n = x n +y n for each n. Determine each of x n and y n explicitly in terms of n. One can express this problem in terms of matrices by writing xn 4 xn u n = = = Au y n y n = u n = A n u. n Let us now focus on computing A n. The characteristic polynomial of A is f(λ) = λ (tra)λ+deta = λ 6λ+9 = (λ 3), so the only eigenvalue is λ = 3. As one can easily check, the corresponding null space is {} N(A 3I) = Span, while (A 3I) is the zero matrix. Thus, a Jordan basis is provided by the vectors v =, v = (A 3I)v =. Letting B be the matrix whose columns are v and v, we must thus have [ 3 3 J = B AB = = J 3] n = B A n n B = n3 n 3 n. Once we now solve this equation for A n, we find that 3 A n = BJ n B n (n+3)3 n n3 = n3 n 3 n = n n3 n (3 n)3 n. In particular, the sequences x n,y n are given explicitly by xn (n+)3 = u n = A n n u = (n )3 n. y n

10 . Which of the following matrices are similar? Explain. A =, B =, C =. Two matrices are similar if and only if they have the same Jordan form. In this case, A is already in Jordan form, all matrices have λ = as their only eigenvalue, while dimn(b I) =, dimn(c I) =. This means that the Jordan form of B consists of one block, so B is similar to A. On the other hand, the Jordan form of C contains two blocks, so C is not similar to either A or B. 3. Let J be a 4 4 Jordan block with eigenvalue λ =. Find the Jordan form of J. Since J is a Jordan block with eigenvalue λ =, it is easy to check that J = = J = = J4 =. In particular, the null space of A = J is two-dimensional and the null space of A = J 4 is four-dimensional, so the Jordan chain diagram is and the Jordan form is J = 4. Use the Cayley-Hamilton theorem to compute A 7 in the case that 3 3 A =. 3 3 The characteristic polynomial of the given matrix is f(λ) = λ 3 λ, so one has. A 3 A = = A 3 = A = A 4 = A 3 = ( ) A. It follows by induction that A n = ( ) n A for each n 3 and this finally gives A 7 = ( ) 5 A = =

11 Linear algebra II Homework #6 solutions. The following matrix has λ = as a triple eigenvalue. Use this fact to find its Jordan form, its minimal polynomial and also its inverse. A = To find the Jordan form of A, we use row reduction on the matrix A λi to get / / A λi = A+I = This gives one pivot and two free variables, so there are two Jordan blocks and J =. The minimal polynomial is m(λ) = (λ+). Since A satisfies this polynomial, one has 3 A +A+I = = I = A(I +A) = A = I A = The following matrix has eigenvalues λ =,,. Use this fact to find its Jordan form, its minimal polynomial and also its power A A =. The eigenvalue λ = is simple, so it contributes an block. When λ =, we have 3 5 A λi = A I = 3 and this gives a single block with eigenvalue λ =. The Jordan form is thus J =

12 and the minimal polynomial is m(λ) = λ(λ ). Since A satisfies this polynomial, A 3 A +A = = A 3 = A A = A 4 = A 3 A = (A A) A = 3A A = A 5 = 3A 3 A = 3(A A) A = 4A 3A. It follows by induction that A n = (n )A (n )A for each n 3 and this gives A 7 = = Let P be the space of all real polynomials of degree at most and let f,g = (3 x) f(x)g(x)dx for all f,g P. Find the matrix of this bilinear form with respect to the standard basis. The standard basis consists of the polynomials v =, v = x, v 3 = x and this gives v i,v j = x i,x j = (3 x) x i+j dx = 3 i+j i+j for all integers i,j 3. Since the matrix of the form has entries a ij = v i,v j, we get 5/ 7/6 3/4 A = 7/6 3/4 /. 3/4 / 3/3 4. Define a bilinear form on R by setting x,y = 4x y +x y +x y +7x y. Find the matrix A of this form with respect to the standard basis and then find the matrix with respect to a basis consisting of eigenvectors of A. The matrix of the bilinear form x,y = i,j a ijx i y j with respect to the standard basis is the matrix A whose (i,j)th entry is the coefficient of x i y j. In our case, we have 4 A =. 7

13 Let us now compute the eigenvectors of this matrix. The characteristic polynomial is f(λ) = λ (tra)λ+deta = λ λ+4 = (λ 3)(λ 8), so the eigenvalues are λ = 3 and λ = 8. The corresponding eigenvectors turn out to be v =, v =. Finally, we find the matrix M with respect to the basis v,v. Since x,y = x t Ay, we get 5 m ij = v i,v j = viav t j = λ j viv t j = λ j (v i v j ) = M =. 4

14 Linear algebra II Homework #7 solutions. Consider R 3 with the usual dot product. Use the Gram-Schmidt procedure to find an orthogonal basis, starting with the vectors v =, v =, 3 v 3 =. Using the Gram-Schmidt procedure, we let w = v and replace the second vector by w = v v,w w,w w = 4 /3 = /3. 3 /3 As for the third vector v 3, this needs to be replaced by w 3 = v 3 v 3,w w,w w v 3,w w,w w =.. Define a bilinear form on R by setting x,y = x y +x y +x y +6x y. Show that this is an inner product and use the Gram-Schmidt procedure to find an orthogonal basis for it, starting with the standard basis of R. The given form is symmetric because its matrix with respect to the standard basis is A =. 6 To show that the form is also positive definite, we complete the square to find that x,x = x +4x x +6x = (x +x ) +x. Finally, one may obtain an orthogonal basis by letting w = e and w = e e,w w,w w = e et Ae e t Ae e = e a a e =.

15 3. Let A = 3. For which values of a is the form x,y = x 4 a t Ay positive definite? Letting y = x and completing the square, one finds that x,x = x +7x x +ax = (x +7x /) +(a 49/4)x. It easily follows that the given form is positive definite if and only if a > 49/4. 4. Let P be the space of all real polynomials of degree at most and let f,g = (4 5x) f(x)g(x)dx for all f,g P. Is this bilinear form positive definite? Hint: compute ax+b,ax+b for all a,b R. Every element of P has the form f(x) = ax+b for some a,b R and this gives f,f = ax+b,ax+b = (4 5x) (ax+b) dx. We now expand the quadratic factor and then integrate to get f,f = = (4 5x) (a x +abx+b )dx ) (4b +(8ab 5b )x+(4a ab)x 5a x 3 dx = 4b + 8ab 5b + 4a ab 3 5a 4. To see that the form is positive definite, it remains to rearrange terms and write f,f = a +8ab+8b = (a+4b) +b.

16 Linear algebra II Homework #8 solutions. Find an orthogonal matrix B such that B t AB is diagonal when 4 A = 3. The eigenvalues of A are the roots of the characteristic polynomial f(λ) = det(a λi) = λ 3 +9λ 8λ = λ(λ 3)(λ 6). This gives λ =, λ = 3 and λ 3 = 6, while the corresponding eigenvectors are v =, v =, v 3 =. Since the eigenvalues are distinct, the eigenvectors are orthogonal to one another. We may thus divide each of them by its length to obtain an orthogonal matrix B such that /3 /3 /3 B = /3 /3 /3 = B t AB = B AB = 3. /3 /3 /3 6. Let P be the space of all real polynomials of degree at most and let f,g = 3x f(x)g(x)dx for all f,g P. Find the matrix A of this bilinear form with respect to the standard basis and then find an orthogonal matrix B such that B t AB is diagonal. By definition, the entries of the matrix A are given by the formula a ij = x i,x j = 3x i+j dx. This gives a ij = when i+j is even and also a ij = 6/(i+j) when i+j is odd, so A =.

17 The eigenvalues of A are λ = and λ =, while the corresponding eigenvectors are v =, v =. Since the eigenvalues are distinct, the eigenvectors are orthogonal to one another. We may thus divide each of them by its length to obtain an orthogonal matrix B such that [ ] / / B = / / = B t AB = B AB =. 3. Show that every eigenvalue λ of a real orthogonal matrix B has absolute value. In other words, show that every eigenvalue λ of B satisfies λλ =. Assuming that v is an eigenvector of B with eigenvalue λ, we get λλ v,v = λv,λv = Bv,Bv = B Bv,v. Since B is real and orthogonal, one has B B = B t B = I n and this implies that λλ v,v = B Bv,v = v,v = λλ =. 4. Suppose that v,v,...,v n form an orthonormal basis of R n and consider the n n matrix A = I n v v t. Show that A is symmetric, orthogonal and diagonalisable. To say that A is symmetric is to say that A t = A and this is true because A t = I t n (v v t )t = I n v tt vt = I n v v t = A. To say that A is orthogonal is to say that A t A = I n and this is true because A t A = AA = (I n v v t )(I n v v t ) = I n 4v v t +4v (v t v )v t = I n. Finally, we show that A is diagonalisable. This follows by the spectral theorem, but it can also be verified directly by showing that each v i is an eigenvector of A. In fact, one has Av = v v (v t v ) = v, Av k = v k v (v t v k ) = v k for each k n. This gives n linearly independent eigenvectors, so A is diagonalisable.