# 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?

## 1 Answer

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
• but eigenvalue of Jordan block can only be 1 as it is upper triangular matrix... ?? – Widawensen Jul 12 '17 at 13:20
• $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ is a Jordan block with eigenvalue $2$. – MooS Jul 12 '17 at 13:22
• I see, I was suggested by the example. So for eigenvalue 2 we have no similarity.. – Widawensen Jul 12 '17 at 13:25