1
$\begingroup$

The exam question was:

$f$ is a linear map where $f^3=f,$ prove that it is diagonizable.

It was a linear algebra exam.

I asked the tutor if $f$ is a matrix since we only defined that a matrix can be diagonizable not any linear map, but he answered that we should prove it for linear maps in general. (Maybe iI missed something in the lectures?)

So $I$ wrote down that $f^3-f=0$ and that we can factor everything out like so: $$f(f+1)(f-1)=0 $$ And argued that this is the minimal polynomial for the function and since its linear factors are different it means that f is diagonizable.

However as far as I know this holds only for matrices? Is this approach correct/applicable for linear maps ? Was my tutor perhaps wrong? Is there something that I am missing?

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Aug 21, 2022 at 21:36

1 Answer 1

1
$\begingroup$

Suppose $f$ is a linear map such that $f^3=f$. Define the following: $$ p_{0}=(1-f^2),\\p_{1}=\frac{1}{2}(f^2+f),\\ p_{-1}=\frac{1}{2}(f^2-f). $$ Note that $p_{0}+p_{1}+p_{-1}=1$. It is not hard to verify that each of these is a projection, meaning that they are idempotent. For example, $$ p_0^2=1-2f^2+f^4=1-2f^2+f^2=1-f^2=p_0,\\ p_1^2=\frac{1}{4}(f^4+2f^3+f^2)=\frac{1}{4}(f^2+2f+f^2)=\frac{1}{2}(f^2+f)=p_1, \\ p_{-1}^2=\frac{1}{4}(f^4-2f^3+f^2)=\frac{1}{4}(f^2-2f+f^2)=\frac{1}{2}(f^2-f)=p_{-1} $$ And these are pairwise disjoint, meaning that $p_jp_k=0$ for $j\ne k$. These are projections onto the eigenspaces of $f$ with eigenvalues $0,1,-1$. Therefore, $$ f = f(p_0+p_1+p_{-1})=p_1-p_{-1} $$ So $f$ is diagonalizable with eigenvalues $0,1,-1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .