<s>[INST] $(\dim_K\ker f)^2 + (\dim_K\operatorname{im} f)^2 = (\dim_K X )^2$ 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?
 [/INST] 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. 
</s>