# The only eigenvalue of $A \in {M_n}$ is $\lambda = 1$. Why is $A$ similar to $A^k$?

Suppose that the only eigenvalue of $A \in {M_n}$ is $\lambda = 1$. Why is $A$ similar to $A^k$ for each $k=1,2,3,\dots$?

• Hint Choose a basis for which $A$ is in Jordan normal form. May 5 '15 at 17:03
• @Elaqqad minimal polynomials are, in general, not sufficient. May 5 '15 at 17:49
• @Underground What have you tried so far? Do you understand Travis' hint? May 5 '15 at 17:54
• @el.Salvador No, $A=\pmatrix{1&1\\0&1}$ and $A^k=\pmatrix{1&k\\0&1}$ are similar obviously.
– daw
May 5 '15 at 19:34

This is not true in general, it depends on the underlying field. Take $A=\pmatrix{ 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 }$. It has only one real eigenvalue (which is one), but is not similar to e.g. $A^4 = I_3$.

In the case of a complex field, we have to show that the Jordan forms of $A$ and $A^k$ are equal. From the construction of the Jordan form it follows that they are equal iff $$rank((\lambda I - A)^n) = rank((\lambda I - A^k)^n)$$ for all eigenvalues $\lambda$ and all powers $n$. Since $\lambda=1$ is the only (complex) eigenvalue it remains to check this case.

Now it holds $$I-A^k = (I-A)(I + A + \dots A^{k-1}).$$ Since $1$ is the only eigenvalue of $A$, $k$ is the only eigenvalue of $I + A + \dots A^{k-1}$, hence this matrix is invertible.

Multiplying with invertible matrices leaves rank invariant, hence $$rank( (I-A^k)^n ) = rank( (I-A)^n(I + A + \dots A^{k-1})^n) = rank( (I-A)^n,$$ which implies that the Jordan forms of $A$ and $A^k$ are equal.

I wonder, whether there is a more direct solution to this question, that is to construct invertible $T$ such that $A=TA^kT^{-1}$.

• if we take $A$ to be in Jordan form, then the similarity $T$ is fairly easy to construct. May 5 '15 at 19:51
• Even if you assume that the eigenvalue $1$ has (algebraic) multiplicity $n$, your first statement is true. For example, when $p$ is a prime and $F$ is the field of $p$ elements, a Jordan block $J_{p}(1)$ of size $p$ with single eigenvalue $1$ has $J_{p}(1)^{p} = I.$ May 5 '15 at 19:54