What are the Eigenvalues of $A^2?$

Question

If $a$ is an eigenvalue of $A$ then $A^2$ has an eigenvalue $a^2.$ But what about the multiplicities? I mean is it possible to have $1,1,2$ as the eigenvalues of a $3\times 3$ matrix $A$ and $1,4,8$ as the eigenvalues of $A^2?$

Answer 1

If $a$ is an eigenvalue of $A$ with multiplicity $k$, then $a^2$ is an eigenvalue of $A^2$ with multiplicity at least $k$. So, the answer is negative: if $1$ is an eigenvalue of $A^2$ and its multiplicity as at least $2$, then $4$ and $8$ can't be eigenvalues too.

Answer 2

Let

$$P(\lambda)=\det(\lambda I-A)=(\lambda-\lambda_1)\cdots(\lambda-\lambda_n)$$

be the characteristic polynomial for $A$ Then

$$\begin{align} \det(\lambda I+A)&=(-1)^n\det(-\lambda I-A)\\ &=(-1)^nP(-\lambda)\\ &=(-1)^n(-\lambda-\lambda_1)\cdots(-\lambda-\lambda_n)\\ &=(\lambda+\lambda_1)\cdots(\lambda+\lambda_n) \end{align}$$

Now if $Q(\lambda)=\det(\lambda I-A^2)$ is the characteristic polynomial for $A^2$, then

$$\begin{align} Q(\lambda^2) &=\det(\lambda^2I-A^2)\\ &=\det(\lambda I-A)\det(\lambda I+A)\\ &=(\lambda-\lambda_1)\cdots(\lambda-\lambda_n)(\lambda+\lambda_1)\cdots(\lambda+\lambda_n)\\ &=(\lambda^2-\lambda_1^2)\cdots(\lambda^2-\lambda_n^2) \end{align}$$

which gives us an identity for the polynomial $Q$. It follows that $Q(\lambda)=(\lambda-\lambda_1^2)\cdots(\lambda-\lambda_n^2)$. Thus the eigenvalues of $A^2$ cannot be anything other than the squares of the eigenvalues of $A$.

Answer 3

If $A$ is a square matrix with eigenvalues $\lambda$, we can apply induction on the size of the matrix to show that the eigenvalues of $A^{k}$ are $\lambda^{k}$ for any positive integer $k$. Hence, the eigenvalues of $A^{2}$ are exactly $\lambda^{2}$ (the squares of the eigenvalues of $A$). See here: Show that $A^k$ has eigenvalues $\lambda^k$ and eigenvectors $v$.