# What are the eigenvalues of a squared matrix?

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 $$A^n$$?

• Note that $A$ might not be diagonalizable if it is larger than $2\times 2$ or if $\lambda_1 = \lambda_2$. Commented Feb 2, 2023 at 16:43
• Suppose $x$ is an eigenvector of $A$ with eigenvalue $\lambda$. Then $A^2x = A(Ax) = A(\lambda x) = \lambda (Ax) = \lambda\cdot \lambda x = \lambda^2 x$. No need to invoke diagonalization arguments Commented Feb 2, 2023 at 16:43
• Terminology note: a square matrix is one with the same number of rows as columns. You might call this a squared matrix. Commented Feb 2, 2023 at 16:43
• @JMoravitz, you are right, thanks. Commented Feb 2, 2023 at 16:46
• @eyeballfrog indeed for the 1st part. But $\lambda_1=\lambda_2$ is not a problem. For example, identity matrix has a single (double) eigenvalue $1$, and is so diagonalizable, that it is diagonal :D Commented Feb 2, 2023 at 17:37

## 1 Answer

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).

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Commented Feb 4, 2023 at 17:55