Math 315: Linear Algebra Solutions to Assignment 7

Transcription

1 Math 5: Linear Algebra s to Assignment 7 # Find the eigenvalues of the following matrices. (a.) (b.) (a.) The characteristic polynomial det(λi A) = (λ )(λ )(λ ), so the eigenvalues are λ =, λ =, and λ =. (b.) The characteristic polynomial det(λi A) = (λ + 8)(λ + ), so the only real eigenvalue is λ = 8. # Find the bases for the eigenspaces of the matrices in Exercise. (a.) For λ =, the reduced row-echelon form of (I A) is , so the general solution of (I A)x = 0 is x = (0, t, 0) = t(0,, 0), so the eigenvector (0,, 0) forms a basis of the corresponding eigenspace. 0 / For λ =, the reduced row-echelon form of (I A) is 0, so the general solution of (I A)x = 0 is x = t( /,, ), and the eigenvector ( /,, ) forms a basis of the corresponding eigenspace.

2 For λ =, the reduced row-echelon form of (I A) is 0 0, so the general solution of (I A)x = 0 is x = ( t, t, t) = t(,, ), so the eigenvector (,, ) forms a basis of the corresponding eigenspace. 0 / (b.) For λ = 8 the reduced row-echelon form of ( 8I A) is 0 /, so the general solution of ( 8I A)x = 0 is x = t( /, /, ), and the eigenvector ( /, /, ) forms a basis of the corresponding eigenspace. # Show that if A is a matrix, then det(λi A) = λ tr(a) λ + det(a), where tr(a) is the trace of A, that is the sum of the numbers on its main diagonal. det(λi A) = λ a a a λ a = (λ a )(λ a ) a a = λ λ(a + a ) + (a a a a ) = λ tr(a) λ + det(a) #4 Prove: If λ is an eigenvalue of an invertible matrix A and x is a corresponding eigenvector, then /λ is an eigenvalue of A, and x is a corresponding eigenvector. Let λ be an eigenvalue of an invertible matrix A. Then λ 0 and Ax = λx for some x 0. The last equality implies A Ax = A λx, which simplifies to x = λa x. This, in turn, can be rewritten as A x = λ x, which means that x is an eigenvector of A to the eigenvalue /λ.

3 #5 Prove that if A is a square matrix, then A and A T have the same eigenvalues. Let E A be the set of eigenvalues of A and E A T be the one for A T. We have to show E E A T. To prove E A E A T, pick λ E A. Then λ is a root of the characteristic polynomial det(λi A) = 0. Transposing the above equation and taking into account that det(b) = det(b T ) for any square matrix B, one has det(λi A) = 0 det(λi A) T = 0 det(λi T A T ) = 0 det(λi A T ) = 0, Which implies that λ is a root of the characteristic polynomial for A T, so λ E A T. The inclusion E A E A T can be proved similarly. # Prove: If A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A. If A is diagonalizable, then there exists an invertible matrix P such that the matrix D = P AP is diagonal. The diagonal of D is formed by the eigenvalues of A. The rank of a diagonal matrix D is obviously equal to the number of its non-zero entries, because the non-zero values of the diagonal correspond to the leading ones in the row-echelon form of D. These non-zero entries correspond to the non-zero eigenvalues of A, so the rank of D is the number of non-zero eigenvalues of A. Problem 0 of Assignment 7 asked to prove that rank(ab) = rank(a) for any invertible matrix B. In a similar way one can also prove that rank(ba) = rank(a) for any invertible matrix B. Note that D = P AP is equivalent to P D = AP. Since P is invertible, applying the above result leads to rank(d) = rank(p D) = rank(ap ) = rank(a). So, rank(d) = rank(a), which is the number of non-zero eigenvalues of A, as we have proved above for the matrix D.

4 #7 Show that if 0 < θ < π, then [ ] cos θ sin θ sin θ cos θ has no eigenvalues. Give a geometric explanation of this result. The characteristic polynomial of A is det(λi A) = (λ cos θ) + sin θ = λ λ cos θ + (cos θ + sin θ = λ λ cos θ +. The roots of this polynomial are given by cos θ ± cos θ. For 0 < θ < π one has cos θ < 0, so the characteristic polynomials has no real roots, which means the matrix A has no eigenvalues. An explanation of this result is that if A is considered as a linear operator in R and x is its eigenvector, then Ax results in stretching x. In our case A is the matrix of a rotation on angle θ, which does not stretch a vector if 0 < θ < π. #8 Find the eigenvalues of the matrix A for n : c c c n c c c n c c c n If λ is an eigenvalue of A, then Ax = λx for a non-zero vector x. In the matrix form the last equation becomes c x + c x + + c n x n c x + c x + + c n x n c x + c x + + c n x n = λ from where we conclude that the vector in the left-hand side of this equation has all the entries equal. Hence, for λ 0 all the entries of the eigenvector x in the right-hand side of the above equation also have to be equal. But then the eigenvector x is a multiple of (,,..., ). This implies c + c + + c n = λ, and this is one of the eigenvalues. The corresponding eigenspace has a basis consisting of a single eigenvector (,,..., ), so the dimension of this eigenspace is, and the multiplicity of the corresponding eigenvalue is also. 4 x x. x n,

5 If λ = 0, then the eigenvector (x, x,..., x n ) has to satisfy the equation c x + c x + + c n x n = 0. This system of linear equations (consisting of a single equation) has leading one. Therefore, the multiplicity of the eigenvalue λ = 0 is the dimension of the null-space of A, which is then n. Since the characteristic equation has at most n roots (counted together with its multiplicities), we have established all the eigenvalues of A: the eigenvalue λ = 0 is of multiplicity n, and the eigenvalue λ = c + c + + c n is of multiplicity. #9 Find A n if n is a positive integer and 0 0 First, let us find a matrix P that diagonalizes A. Since A is symmetric and has different eigenvalues λ =, λ =, and λ = 4, the diagonalizing matrix P should be orthogonal, so D = P T AP and P DP T. The unit length eigenvectors corresponding to the above eigenvalue are, respectively, (,, ), (, 0, ), and (,, ). Hence, Finally, A n = P D n P T = = = n 4 n 0 4n n 4 n ) ( ( n + 4n ( ) ( 4n ( n + ) ( 4n n n ) ( + n + 4n + 4n ) ( 4n ) 4n + 4n ) ( n + 4n ) ) 5

6 #0 Find a matrix P that diagonalizes A, and determine P AP for The characteristic polynomial of A is det(λi A) = (λ )(λ )(λ ). Hence, the diagonalized matrix D = P AP is D = The eigenvectors corresponding to the eigenvalues λ =, λ =, and λ = are, respectively, (,, ), (,, ), and (,, 4). So, the diagonalizing matrix P is P = 4