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

  • 5
    $\begingroup$ 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

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 $A=TA^kT^{-1}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.