Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

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!

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.