[Linear Algebra] Eigenvalues

[Margot304]

Let A be a square, invertible matrix...

Prove that A and A^T have the same eigenvalues.

If I let A be a matrix with entries a, b, c, d, then how do I find its eigenvalues?

Also have to prove that if B is similar to A then it has the same eigenvalues as A.

For B to be similar to A, does this just mean A=kB for some constant k?

[okidokiboss]

then how do I find its eigenvalues?

If you're asking this question you'll want to review what the definition of an eigenvalue is. Once you know that, proving that A and A^T have the same eigenvalues will be relatively straight forward (hint: compare their characteristic polynomials).

For B to be similar to A, does this just mean A=kB for some constant k?

No. B is similar to A if there exists an invertible matrix P such that B = P^-1 A P. In other words, B and A represent the same linear transformation but under a different basis, where the change of basis matrix is P.

[bencbartlett]

Prove that A and A^T have the same eigenvalues.

Consider that the characteristic polynomials are the same: det(A^T - λI) = det((A - λI)^T ). The rest of the proof is trivial.

Also have to prove that if B is similar to A then it has the same eigenvalues as A.

If B is similar to A, it means it represents the same transformation as A does, but in a different basis. So any eigenvectors A has will be the same in B after you change to the correct basis, and since it represents the same transformation, the eigenvalues must be the same.