Show that $f\big(\text{Ker}(g)\big)\subseteq \text{Ker}(g)$ and $f\big(\text{Im}(g)\big)\subseteq \text{Im}(g)$.

Question

Let $f, g:V\to V$ be linear maps on a vector spce $V$ such that $f\big(g(v)\big)=g\big(f(v)\big)$ for all $v\in V$.

1) Is is true that $f\big(\text{Ker}(g)\big)\subseteq \text{Ker}(g)$?

2) Is it true that $f\big(\text{Im}(g)\big)\subseteq \text{Im}(g)$?

What does it mean $f\big(\text{Ker}(g)\big)$? Does it mean the image of $\text{Ker}(g)$? I have no idea how to start these kind of questions....

Answer 1

I assume that $f$ and $g$ are linear maps. The meanings of $f(\ker g)$ and $f(\operatorname{im} g)$ are, respectively, the image of $\ker g$ and the image of $\operatorname{im} g$ under $f$, i.e., $$f(\ker g)=\big\{f(x):x\in \ker g\big\}$$ and $$f(\operatorname{im} g)=\big\{f(y):y\in \operatorname{im} g\big\}.$$

(1) Let $x\in \ker g$. Then, $$g\big(f(x)\big)=f\big(g(x)\big)=f(0)=0,$$ so $f(x)\in \ker g$. This proves that $f(\ker g)\subseteq \ker g$.
More generally, if $V_\lambda(g)$ is the eigenspace of $g$ with eigenvalue $\lambda$, then $f\big(V_\lambda(g)\big)\subseteq V_\lambda(g)$. Note that $\ker g= V_0(g)$. (We also have $f\big(V^\lambda(g)\big)\subseteq V^\lambda(g)$ if $V^\lambda(g)$ is the generalized eigenspace of $g$ with eigenvalue $\lambda$.)

(2) Let $y\in \operatorname{im}g$. Then, $y=g(x)$ for some $x\in V$. Therefore, $$f(y)=f\big(g(x)\big)=g\big(f(x)\big)\in \operatorname{im}g.$$ Hence, $f(\operatorname{im}g)\subseteq \operatorname{im}g$.

Answer 2

Given any set $A\subset V$, $$ f(A):=\{f(a):a\in V\}. $$ So, you can think about this as the image of $f$, when restricted to the set $A$, as you mentioned.

For part 1, then, $$ f(\ker g)=\{f(v):v\in \ker g\}. $$ Given $w\in f(\ker g)$, we know by the above that there is a $v$ in $\ker g$ such that $w=f(v)$. So, now the question is "Is $w$ also in $\ker g$?" I.e., is $g(w)=g(f(v))=0$? You know is that $g(v)=0$. You also know that $f(g(v))=g(f(v))$.