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

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

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  • $\begingroup$ if we take $A$ to be in Jordan form, then the similarity $T$ is fairly easy to construct. $\endgroup$ May 5 '15 at 19:51
  • $\begingroup$ 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 $J_{p}(1)^{p} = I.$ $\endgroup$ May 5 '15 at 19:54

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