Basis of vector space such that the matrix of a linear transformation with respect to the basis has a desired block form 
Let $V$ be an $n$-dimensional vector space over a field $F$. Let $T : V \to V$ be a linear transformation. Describe bases $B$ and $B'$ of $V$ such that the matrix of $T$ with respect to $B$ and $B'$ has the form
$$\left[\begin{array}{c|c} I_k & 0 \\ \hline 0 & 0 \end{array}\right]$$
for some $k \leq n$. Deduce that if $A$ is an $n \times n$ matrix over $F$, then there exist invertible $n \times n$ matrices $P$ and $Q$ such that $PAQ$ has the block form above.

Here is my thinking so far:
Let $B = \{v_1,.v_2,...,v_n\}$ be an ordered basis of $V$ such that $T(v_i) = v_i$ for $i \leq k$ and $T(v_j) = 0$ for $k < j \leq n$. Then the matrix of $T$ with respect to $V$ will have the desired block form above.
However, I don't see how to get another ordered basis $B'$ of $V$ with the same block form. Does this amount to a linear transformation with the same idea, except including some scalars in the field $F$ ?
Also, I'm not quite how to deduce we have invertible matrices $P$ and $Q$ such that $PAQ$ has the desired block form. Does this have to do with the formula for a change of basis matrix ? Or does it have to do with observing that the desired block form is in Jordan form, with $k$ $1 \times 1$ blocks corresponding to eigenvalue $1$ and $n-k$ $1 \times 1$ blocks corresponding to eigenvalue $0$ ?
Thanks!
 A: Arin's answer is already very clear. I hope my comment makes it more intuitive.
Basically, the issue with your argument is that the $B=\{v_1,v_2,...,v_n\}$ that you were trying to find may not exist.
If $B=\{v_j\}$ is a basis for the vector space you're mapping from, and $B'=\{w_i\}$ is a basis for the vector space you're mapping onto, then the matrix $A=(a_{ij})$ representing the transformation $T$ satisfies that $T(v_j)=a_{ij}w_i$.
If you have a set $\{v_i\}$ such that $T(v_i)=v_i$ for $i \leq k$ and $T(v_j)=0$ for $j>k$, what you have is indeed a matrix $A$ of the desired form. But note that in this case, the bases you choose for the domain space and the image space are the same, which is not necessary.
If you could see that $T(v_j)=a_{ij}w_i$, then all you need to do is to find a set of $v_j$ such that $\{T(v_j)\}$ forms a basis for the image. Then, let $w_j=\{T(v_j)\}$ and you get the desired form of the matrix.
And you get the invertible matrices $P$ and $Q$ in the problem by the two "change of basis" operations.
A: Hint:
Let $r$ be the rank of $T$, then $q = n - r$ is the dimension of null space of $T$ where $n = \text{dim}(V)$.
Let $x_1,x_2, \dots, x_n$ be a basis of $V$ such that $x_{r+1},\dots,x_{n}$ is a basis of null space of $T$.
Let $y_1 = T(x_1), \dots, y_r = T(x_r)$.   Show that $y_1,y_2,\dots,y_r$ are linearly independent.  So you can extend it to the basis of $V$ say, $y_1,y_2,\dots,y_r,y_{r+1},\dots,y_n$.
Then $x_1,\dots,x_n$ and $y_1,y_2,\dots,y_n$ are your required basis.
