Suppose that $V$ is a finite dimensional $\mathbb C$-vector space, and suppose that $T:V\rightarrow V$ is injective. If there is a $m\in\mathbb N$ such $T^m$ is diagonalizable, then $T$ is diagonalizable.

I've found the proof for the case that $T^m=Id$, but I can't adapt it to this case.

  • $\begingroup$ Notice that $T$ is injective if and only if all of it's eigenvalue is non-zero. Then try expressing $T^m$ in terms of the jordan matrix of $T$. $\endgroup$ – user341124 Jul 7 '17 at 17:17
  • $\begingroup$ Fixed horrible problem with my answer. $\endgroup$ – hmakholm left over Monica Jul 7 '17 at 17:46

Choose a basis for $V$ such that the matrix of $T$ is in Jordan normal form.

If the matrix is not diagonalizable, then its Jordan normal form has one or more Jordan blocks of size $\ge 2$. Its $m$th power is a block diagonal matrix where each block is the $m$th power of a Jordan block.

A block diagonal matrix is diagonalizable iff each of the blocks are. However the power of a non-trivial Jordan block is a matrix whose eigenvalues are all equal, and such a matrix is diagonalizable if and only if it is already diagonal.

The power of a Jordan block can be found using the rule here, and when $\lambda$ is nonzero (which it is because $T$ is injective), we can see that the powers are never diagonal (unless the block is 1×1, of course).

  • $\begingroup$ One idea: you can separate your expression for $T^m$ into $D + N$ where $D$ and $N$ are commuting, $D$ is diagonalizable (in fact diagonal) and $N$ is nilpotent. $\endgroup$ – Ben Grossmann Jul 7 '17 at 17:38
  • 2
    $\begingroup$ @Omnomnomnom: I think I made it work with my original reasoning. $\endgroup$ – hmakholm left over Monica Jul 7 '17 at 17:48

Proof: note that a transformation $M$ is diagonalizable if and only if there exist distinct $\lambda_i \in \Bbb C$ such that $$ \prod_{i=1}^k (M - \lambda_i I) = 0 $$ So, select distinct $\lambda_i \neq 0$ such that $$ \prod_{i=1}^k (T^m - \lambda_i I) = 0 $$ Now, let $\mu_{i1},\dots,\mu_{im}$ denote the solutions to $z^m = \lambda_i$. We can write $$ 0 = \prod_{i=1}^k (T^m - \lambda_i I) = \prod_{i=1}^k \left(\prod_{j=1}^m (T - \mu_{ij} I)\right) = \prod_{i=1, j=1}^{k,m} (T - \mu_{ij} I) $$ which is a product of distinct linear factors. So, $T$ is diagonalizable.


We use:

A matrix $M$ is diagonalizable if and only if the minimal polynomial for $M$ has no repeated roots.


Give a matrix $M$ and a polynomial $p$ such that $p(A)=0,$ then $p(x)$ is divisible by the minimal polynomial of $M$.

Now, let $m_A(x)$ be the minimal polynomial for a matrix $A.$

If $m_{T^m}(x)$ has no repeated roots, then $p(x)=m_{T^m}(x^m)$ can only have $x=0$ as a repeated root. (I'll leave that to you to prove [*].)

But $p(A)=M_{T^m}(T^m)=0$, so $p(x)$ is divisible by the minimal polynomial, $m_{A}(x),$ of $A.$ If $A$ is non-singular, then $0$ is not a root of $m_{A}(x).$ So the minimal polynomial for $A$ has no repeated roots.

For [*] you'll need to prove this in some form:

Over an algebraically closed field $k$ of characteristic $0,$ (such as $\mathbb C,)$ there are no repeated roots to $x^n-1.$

From there you can conclude that:

For $a\in k,$ $x^n-a$ has repeated roots if and only if $a=0.$


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.