

Proof. We have that
so any solution of det(A –λI) = 0 is a solution of det(A –λI)T = 0 and vice versa. Thus A and AT have the same eigenvalues. 

The matrices A and AT will usually have different eigenvectors. 

Proof. Suppose the matrix A is diagonal or triangular. If you subtract λ's from its diagonal elements, the result A – λI is still diagonal or triangular. Its determinant is the product of its diagonal elements, so it is just the product of factors of the form (diagonal element – λ). The roots of the characteristic equation must then be the diagonal elements. 
Another addition to the square matrix theorem.

Proof. An n x n matrix A has an eigenvalue 0 if and only if det(A – 0I) = 0, i.e. if and only if det(A) = 0. Since A is invertible if and only if detA ≠ 0, A is invertible if and only if 0 is not an eigenvalue of A. 

Proof. Suppose the matrix A has eigenvalue λ with eigenvector x, i.e. suppose that Ax = λx. Then A2x = A(Ax) = A(λx) = λ(Ax) = λ(λx) = λ2x. Multiply by more A's to get A3x = λ3x, A4x = λ4x and so on. 

Proof.
Multiply the equation Ax
= λx by λ–1A–1: 


Proof. Suppose the statement is not true, i.e. suppose that A has a linearly dependent set of eigenvectors each with a different eigenvalue. "Thin out" this set of vectors to get a linearly independent subset v1, v2, ..., vk, with distinct eigenvalues λ1, λ2, ..., λk. 

Suppose u
is one of the eigenvectors you thinned out because it was linearly dependent
on the others: 

First multiply * by A:
Since u is also an eigenvector, Au
= λu for some eigenvalue λ,
so this equation gives 

Now multiply * by λ: 

Subtract *** from ** to get
Since the vi's are linearly independent, ci(λi – λ) = 0 for all i = 1, 2, ..., k. Since the eigenvalues (including λ) are all different, ci = 0 for all i. But this implies (from equation *) that u = 0, which is impossible since u is an eigenvector. 

The original assumption must be false, i.e. it is not possible to have a linearly dependent set of eigenvectors with distinct eigenvalues; any eigenvectors with distinct eigenvalues must be linearly independent. 