What are the eigenvalues of a squared matrix?

Question

Suppose matrix $A$ has eigenvalues $\lambda_1$ and $\lambda_2$.
Are the eigenvalues of $A^2$: $\lambda_1^2$ and $\lambda_2^2$?
If so, can I prove this by simple diagnolization, where $T$ is the eigenbasis and $T^{-1}$ its inverse and simply because $T^{-1}A^2T$ equals the identity matrix stretched by corresponding squared eigenvalues so eigenvalues of $A^2$ are squared eigenvalues of $A$?
Also, can this be generalized into $A^n$?

Answer 1

Counter example. In ℝ-space ℝ⁴, Consider the matrix

$$ A = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 \end{pmatrix} $$

It has two eigenvalues, $\lambda_1=1$ and $\lambda_2=2$.

Now compute $A^2$ $$ A^2 = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 4 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix} $$

Which has three eigenvalues, $\mu_1 = \lambda_1^2 = 1$, $\mu_2 = \lambda_2^2 = 4$ and $\mu_3=-1$.

So, no, eigenvalues of $A^2$ are not just the squares of eigenvalues of $A$.

(But the squares of eigenvalues of $A$ are all eigenvalues of $A^2$. If $\lambda$ is an eigenvalue of $A$, then $\exists u, Au=\lambda u$, and then $A^2 = A.Au = A\lambda_u = \lambda Au = \lambda \lambda u = \lambda^2 u$. But that is not the same result as "eigenvalues of $A^2$ are the eigenvalues of $A$, squared")

As for your proof, it contains the words "eigenbasis". It is therefore valid only if there is an eigenbasis. It is only the case if $A$ is diagonalizable (which, by definition, means that it exists a basis made of eigenvectors). My counter example was carefully so that it is not diagonalizable (it is just a combination of a diagonal, and of a classical $\begin{pmatrix} 0 & -1\\ 1 &0\end{pmatrix}$ antipattern for diagonalization).