# 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 $$dim(\ker(f)) = 0$$?

• $f(x)=0$ implies $f^{2}(x)=0$. Jan 4 at 12:12
• Why would the kernel of a linear map be empty? :( Jan 4 at 12:13
• @KaviRamaMurthy nowhere is stated that $f(x) = 0$.
– agus
Jan 4 at 12:18
• @AlvinL that's why I added the clarification. In my university they teach that if the only vector in the Kernel is zero, then we can say the Kernel is empty. To avoid this kind of doubt I edited the question.
– agus
Jan 4 at 12:18
• $x \in Ker (f)$ implies $f(x)=0$, by definition. Jan 4 at 12:21

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 $$\text{Ker}f=\{0\}$$.