Question

Assume all matricies are Mmxm(R) unless otherwise specified. 1. (1 point) Prove or disprove that the eigenvalues of A and AT are the same. 2. (2 points) Let A be a matrix with m distinct, non-zero, eigenvalues. Prove that the eigenvectors of A are linearly independent and span R”. Note that this means in this case) that the eigenvectors are distinct and form a base of the space. 3. (1 point) Given that is an eigenvalue of A associated with v, prove that v is an eigenvector of Ak and that lk is it's associated eigenvalue. 4. (2 points) Let A be a matrix with eigenvalues lk, 1 <k <m. Show that det A = 11 12 ... Im. 5. (1 point) Prove that a matrix A is singular if and only if it has a zero eigenvalue (hint: use the previous problem).

Answer 1

1. The eigenvalues of a square matrix are given by the roots of it's characteristic polynomial. So, the characteristic polynomial of A transpose is given by

$\small \det(A^T-\lambda I)$

Since the transpose of the identity matrix is the identity matrix itself, so we can write

$\small = \det(A^T-\lambda I^T)$

$\small = \det((A-\lambda I)^T)$

Now, taking the transpose of a matrix does not change the determinant, so

$\small = \det((A-\lambda I)^T)=\det(A-\lambda I)$

which is the expression for the characteristic polynomial of A transpose.

Thus, A and A transpose have the same characteristic polynomial, hence the roots of these characteristic polynomial will be the same as well - which means the eigenvalues of the two matrices are the same.

thus our proof is complete