$(\dim_K\ker f)^2 + (\dim_K\operatorname{im} f)^2 = (\dim_K X )^2$

Question

So I have $X$, $Y$ vector spaces over a field $K$ and $f : X\to Y$ linear operator.

The problem asks me to define $\ker f$. Next, it says that given

$(\dim_K\ker f)^2 + (\dim_K\operatorname{im} f)^2 = (\dim_K X)^2$ and $\dim_K\ker f \neq 0$

I must show that $\dim_K\operatorname{im} f = 0$

Any ideas?

Answer 1

By the Rank-Nullility Theorem: $$\dim_K \ker f + \dim_K \operatorname{im} f = \dim_K X\tag{1}$$ Now $$(\dim_K\ker f)^2 + (\dim_K\operatorname{im} f)^2 = (\dim_K X)^2\tag{2}$$ and $\dim_K\ker f \neq 0$. Leting $\dim_K \ker f=a$, $\dim_K \operatorname{im} f =b$, and $\dim_K X=c$, for some $a$, $b$, $c\in\Bbb{Z}_{\ge0}$then by $(1)$, $$a+b=c\tag{3}$$ But by $(2)$, $$a^2+b^2=c^2\tag{4}$$

The question then is: is $(4)$ possible in light of $(3)$ with $b\neq0$.

By $(3)$ you get $$a^2=c^2-2bc+b^2\tag{5}$$ By $(4)$ you get $$a^2=c^2-b^2\tag{6}$$ So by $(5)$ and $(6)$ $$c^2-2bc+b^2=c^2-b^2\implies b^2=bc$$ or $b=c$, which is impossible by $(3)$ as $\dim_K\ker f =a>0$, and so we must have $\dim_K \operatorname{im} f=b=0$ as required.

Answer 2

Since by rank-nullity theorem

dim(ker $f)+$dim(im $f$) $=$ dim($X)$.

Squaring both sides yields

dim(ker $f$) $\cdot$ dim(im $f)=0$,

but dim(ker $f$) is nonzero so dim(im $f) =0.$