Prove the following Let A be a square matrix Show that A and

Prove the following. Let A be a square matrix. Show that A and A^T have the same eigenvalues. A square matrix A is nilpotent if A^k = 0 for some positive integer k. Show that a nilpotent matrix A cannot have nonzero eigenvalues. Let A be an m times n matrix. Show that if A^T A has eigenvalue lambda notequalto 0. then lambda is also an eigenvalue of AA^T.

Solution

The matrix (AI)T is the same as the matrix (ATI), since the identity matric is symmetric.

Thus:

det(ATI)=det((AI)T)=det(AI)

From this it is obvious that the eigenvalues are the same for both A and AT

If vv is a non-zero eigen vector corresponding to an eigenvalues we have, by definition, Av=v. Then A2v=A(Av)=A(v)=2v. It easily follows that n is an eigenvalue for AnAn but the latter is the zero matrix, for which all eigenvalues are zero, hence =0.

(c)

0 is an eigenvalue of ATAdet(ATAI)=0det(I+(1/)ATA)=0det(I+A(1/)AT)=0det(AATI)=00 is an eigenvalue of AAT

The matrix (AI)T is the same as the matrix (ATI), since the identity matric is symmetric. Thus: det(ATI)=det((AI)T)=det(AI) From this it is obvious that the eigenvalues are the same for both A and AT (b) If vv is a non-zero eigen vector corresponding to an eigenvalues we have, by definition, Av=v. Then A2v=A(Av)=A(v)=2v. It easily follows that n is an eigenvalue for AnAn but the latter is the zero matrix, for which all eigenvalues are zero, hence =0. (c) 0 is an eigenvalue of ATAdet(ATAI)=0det(I+(1/)ATA)=0det(I+A(1/)AT)=0det(AATI)=00 is an eigenvalue of AAT