Effect of squaring and the identity-matrix on eigenvalues and the characteristic polynomial

Question

I am new to Linear Algebra, and would like some feedback regarding the following question:

True or false?

Let $A$ be a square matrix over $R$

1. If 3 is an eigenvalue of $A$, then 10 is an eigenvalue of $A^2+I$.
2. If P(t) is the characteristic polynomial of $A$, then $P(t^2)$ is a characteristic polynomial of $A^2$.

I worked out 1. to be true, but simply by building some examples and seeing how squaring and adding $I$ affects the eigenvalues.

I believe 2. is false, but I did this by just trying some numbers. I would love to know if I am right, and what the theory behind it is.

Thank you!

Answer 1

For 1, let $v$ be an eigenvector relative to $3$, that is, $Av=3v$. Then $$ (A^2+I)v=A(Av)+v=A(3v)+v=3Av+v=9v+v=10v $$

The characteristic polynomial of $A$ is $p(t)=\det(A-tI)$, which has degree $n$ (if $A$ is $n\times n$).

Therefore $p(t^2)$ cannot be the characteristic polynomial of $A^2$, because it has degree $2n$.