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Please can you help me whit this problem. For $1.$ I did it as it's classical. What I am having trouble with are the other questions. Hints would be good but if you can explain that would be great.

Let $E, E'$ be two vector spaces over $\mathbb{K}$, $\phi$ is a linear map from $E$ to $E'$ and $F$ (resp $F'$) a subspace of $E$ (resp of $E'$).

$1.$ $a)$ Show that $\phi(F)$ (resp $\phi^{-1}(F')$) is a subspace of $E'$ (resp of $E$).

$b)$ deduce that $Ker\phi$ (resp $\text{Im}\phi$) is a subspace of $E$ (resp of $E'$).

$2.$ we suppose that the dimension of $E$ is finite. Let $f, g$ be two endomorphisms of $E$ and $V$ be a subspace of $E$.

$(i)$ Show that $$\dim f^{-1}(V)= \dim(\text{Ker}f)+\dim(V\cap \text{Im}f)$$

$(ii)$ Deduce that $$\dim (\text{Ker}f\circ g)\leq \dim(\text{Ker}g)+\dim(\text{Ker}f)$$

$(iii)$ Show that $$\text{Ker}(g\circ f)= f^{-1}(\text{Ker}g\cap \text{Im}f)$$

$3.$ We suppose that $f\circ g=g\circ f$.

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

$4.$ We suppose that the dimension of $E$ is finite.

Show that $$\dim(\text{Im}f\cap \text{Ker}g)=\text{rank}f-\text{rank}(g\circ f)$$

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up vote 1 down vote accepted

Since 1 is done, I will try the rest.

For 2 (i), try to use the rank-nullity theorem.

For 2(ii), apply the formula to $g^{-1}(\ker f)$ and use $\dim_{K}(M+N)=\dim(M)+\dim(N)-\dim(M\cap N)$.

For 2(iii), $v$ lies in the kernel iff $g\circ f(v)=0$. So $f(v)$ lies in the kernel of $g$ and in the image.

For 3, use the definition of the kernel, image and commutativity.

For 4, use $\mbox{rank}(g\circ f)+\dim(\ker g|_{f(V)})=\mbox{rank}(f)$. (rank-nullity theorem again)

Hope these helps!

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