Is it true that dim(Ker f) = 0 if f is an endomorphism in R^6, st f^2 = -Id?

I came up with this in an exam but I can't solve it. I don't know about eigenvectors yet, so can someone solve this without them?

Let $f$ be an endomorphism in $\mathbb{R}^6$, such that $f^2 = -Id_{6}$. Is it true that $ dim(\ker(f)) = 0$?

Thanks in advance.

Solution 1:

First of all notice that $\text{Ker}f$ can't be the empty set beacuse always $0\in\text{Ker}f$, so you should write $\text{Ker}f=\{0\}$.

Verify that $\{0\}\subset\text{Ker}f\subset\text{Ker}f^2$, because if $x\in\text{Ker}f$, then $f^2(x)=f(f(x))=f(0)=0$ and so $x\in\text{Ker}f^2$.

But if $f^2=-I$ then $\text{Ker}f^2=\{0\}$. This implies $\text{Ker}f=\{0\}$.