how many linear transformation like $T$ are from $V_F$ to $V_F$ such that $W=\ imT$ or$W=\ kerT$? ($V_F$and $F$ is finite ) let $V_F$ be finite vector space and $W$ is subspace of $V_F$ and $F$ is finite  now 
how many linear transformation  like $T$ are from $V_F$ to $V_F$( $T:V_F \to V_F)$ such that $W=\ imT$ and how many linear transformation  like T are from $V_F$ to $V_F$( $T:V_F \to V_F)
$ such that $W=\ kerT$? 
Thanks in advance 
 A: Pick a basis for $V_F$. Then $T$ is completely determined by where it sends its basis vectors.
If the dimension of $V_F$ is $n$ and the dimension of $W$ is $m$, pick a basis for $V_F$ including a basis for $W$. Then for the kernel of $T$ to be $W$, it needs to send those $m$ basis vectors to $0$ and the other $n - m$ vectors to a linearly independent set of vectors.
Say $n - m = k$, so we want to find how many ways there are to pick vectors $v_1,v_2,\ldots,v_k$ that are linearly independent. If the size of $F$ is $q$, then there are $q^n - 1$ choices for $v_1$ (anything but the origin). Then, the span of $v_1$ consists of $q$ points, so $v_2$ can be any of $q^n - q$ vectors. Similarly $v_3$ can be any of $q^n - q^2$. In general, we get that the total number of linear transformations: having $W$ as a kernel:
$$\prod_{i = 0}^{k - 1} (q^n - q^i)$$
For $W$ as the image, we do two things: First, we have to choose a $k-$dimensional subspace to be the kernel. We do this in essentially the same way: We pick $v_1,v_2,\ldots,v_k$ linearly independent, to be the basis of our subspace. Then we divide by the number of isomorphisms of our subspace. We can count this as simply the number of maps from this space to itself with empty kernel, so the number of $k-$dimensional subspaces is:
$$\frac{\prod_{i = 0}^{k - 1}(q^n - q^i)}{\prod_{i = 0}^{k - 1}(q^k - q^i)}$$
Once we have our kernel picked out, pick a basis for the remainder and map those to some basis of $W$. This can be done in the same number of ways as there are isomorphisms of $W$, so the final number of maps with $W$ as an image:
$$\frac{\prod_{i = 0}^{k - 1}(q^n - q^i)\prod_{i = 0}^{m - 1}(q^m - q^i)}{\prod_{i = 0}^{k - 1}(q^k - q^i)}$$
A small note on these formulas: They don't make a ton of sense when $m = 0$ or $m = n$, when you should replace a product from $i = 0$ to $-1$ with the number $1$. 
A: Hint: Consider scalar multiples.
A: Let $n:=\dim V$ and $m:=\dim W$, and $q:=|F|$. Linear transformations $V\to V$ correspond to $n\times n$ matrices, wrt. a fixed basis. For 2, it is convenient to use a basis of $V$ that extends a basis of $W$.
For 1., let $\def\im{\rm im\,} \mathcal U(W):=\{T\,\mid\,{\rm im\,}T\subseteq W\}$, and let $\mathcal H(W):=\{T\,\mid\,\im T=W\}$. Then we can count $|\mathcal U(W)|$ quite simply: each column of the matrix has to be in $W$, so there are $(q^m)^n$ such matrices. Then
$$|\mathcal H(W)|=|\mathcal U(W)|-\sum_{\matrix{W'<W}}|\mathcal H(W')|\,.$$
So, we are left to count how many subspaces $W$ has of each dimensions $<m$, and we can proceed inductively, starting with $m=0$ then $m=1$...
Similar approach can work for question 2., by considering $\mathcal K(W):=\{T\,\mid\,\ker T\supseteq W\}$.
