<s>[INST] finding the characteristic polynomial of $A$ given the characteristic polynomial of $A^2$ $A$ is a $3\times3$ matrix, the Jordan form of $A^2$ is $$
\begin{pmatrix}
25 & 1 & 0\\
0 & 25 & 0\\
0 & 0 & 1\\
\end{pmatrix}
$$
Find the characteristic polynomial, minimal polynomial and Jordan form of $A$ if $A$ has only negative eigenvalues.
I so far got to:
$P_{A^2}(X)=(x-25)^2(x-1)$, by cayley hamilton:
$P_{A^2}(A^2)=(A^2-25I)^2(A^2-I)=[(A-5I)(A+5I)]^2(A-1)(A+1)=0$
Therefore, $[(A-5I)(A+5I)]^2(A-1)(A+1)$ is divided by the minimal polynomial of $A$, and given all the eigenvalues of $A$ are negative meaning it is one of: $(x+1), (x+5), (x+5)(x+1), (x+5)^2(x+1), (x+5)^2$.
If $A$ is a diagonalizable matrix there is a matrix $M$ so $D=M^{-1}AM$ and $D$ is a diagonal matrix, however, by this equation we also get $D^2=M^{-1}A^2M$, of course $D^2$ is also diagonal meaning $A^2$ is diagonalizable in contradiction to the fact the Jordan form of $A^2$ is not diagonalizable.
Therefore, the minimal polynomial of $A$ is not a product of distinct linear factors, leaving us with $(x+1), (x+5), (x+5)^2(x+1), (x+5)^2$.
I can't figure out how to eliminate the other options, I know the first two options and the last one are incorrect because if the only eigenvalue of $A$ is $-1$ then the $25$ can't be an eigenvalue of $A^2$, I just don't know how to prove it (there is no proof in our book in relation to eigenvalues and power of matrices). Thanks.
 [/INST] Let $\lambda_1, \lambda_2, \lambda_3$ the eigen values of $A$. $$P_A(X) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$$
Since the eigen values of $A$ are $25$, $25$, $1$. So you can assume that $\lambda_1^2=\lambda_2^2=25$ and $\lambda_3^2=1$. You can find the values of $\lambda_i$ since they are negative.
</s>