# Prove that $A$ is similar to $A^n$ based on A's Jordan form

Let $A = \begin{bmatrix}1&-3&0&3\\-2&-6&0&13\\0&-3&1&3\\-1&-4&0&8\end{bmatrix}$, Prove that $A$ is similar to $A^n$ for each $n>0$.

I found that the characteristic polynomial of $A$ is $(t-1)^4$, and the minimal polynomial is $(t-1)^3$. And the Jordan form of $A$ is \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&1\end{bmatrix}

I guess the key to solve this is to use the fact that two matrices are similar if and only if they have the same Jordan form.

Any hints?

After computing the Jordan form you only have to check that all powers of $$M=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$$ are similiar to each other.
Note that $M=I+N$ where $N^3=0$, so $$M^n = I+nN+\frac{n(n-1)}{2}N^2.$$
We obtain $(M^n-I)^2=n^2N^2 \neq 0$, in particular the minimal polynomial of $M^n$ is $(t-1)^3$, i.e. the Jordan form of $M^n$ is precisely $M$.
• Can we generalize the statement "the Jordan form of $M^n$ is precisely $M$" for Jordan block $M$ of any dimension ? – Widawensen Jul 12 '17 at 13:14
• If the eigenvalue of the block is $1$, this should be true for all dimensions. – MooS Jul 12 '17 at 13:18
• $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ is a Jordan block with eigenvalue $2$. – MooS Jul 12 '17 at 13:22