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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.

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  • $\begingroup$ 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
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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).

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  • $\begingroup$ 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
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    $\begingroup$ @Omnomnomnom: I think I made it work with my original reasoning. $\endgroup$ – hmakholm left over Monica Jul 7 '17 at 17:48
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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.

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We use:

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

Also:

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 $a=0.$

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