Rank of linear operator on $M_{3\times 3}(R)$ The question asks the rank of linear operator $T$ on $M_{3 \times 3}(R)$ defined as: 

$T(P)=QP$

where Q is a square matrix of order 3 whose rank is 2.
I can't proceed with this as I have only come across operators on Euclidean n-dimensional space.
 A: One definition of rank (of a linear map) is the dimension of the image.
Thus consider the linear map $T:M \rightarrow M$, where $M$ stands for $M_{3\times 3}(\Bbb R)$. Since we know $\dim M = 9$, to get the dimension of its image, it suffices to calculate the dimension of its kernel.
The map $T$ is defined by $T(P) = QP$, so $\ker(T) = \{P\in M: QP = 0\}$. If we write $P$ as $(P_1, P_2, P_3)$, with $P_i$ column vectors, then $P$ is in $\ker(T)$ if and only if $QP_i = 0$ for $i = 1, 2, 3$.
But $Q$ is a matrix of rank $2$, hence $QP_i = 0$ is equivalent to saying that $P_i$ is in the one dimensional kernel of $Q$ in $\Bbb R^3$.
Hence we conclude that $\dim \ker(T) = 3$, and thus the rank of $T$ is $9 - 3 = 6$.
A: Method $1$: Direct calculation.
Suppose, $Q=\begin{bmatrix}a&0&0\\0&b&0\\0&0&0\end{bmatrix}$ with $a,b\in \Bbb R\setminus\{0\}$, then $Q$ is a $3\times 3$ matrix of rank $2$.
Given $T:M_{3\times 3}(\Bbb R)\to M_{3\times 3}(\Bbb R)$ by $T(P)=QP,\forall P\in M_{3\times 3}(\Bbb R)$. We find the matrix of $T$ w.r.t. canonical basis of $M_{2\times 2}(\Bbb R)$ which is $\{e_{ij}:1\le i\le 3,1\le j\le 3\text{, $e_{ij}$ is a matrix with $(i,j)^{th}$ element as $1$ and rest is zero}\}$
Observe that, $$\begin{bmatrix}a&0&0\\0&b&0\\0&0&0\end{bmatrix}\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}=\begin{bmatrix}a&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
$$\begin{bmatrix}a&0&0\\0&b&0\\0&0&0\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}=\begin{bmatrix}0&a&0\\0&0&0\\0&0&0\end{bmatrix}$$
$$\begin{bmatrix}a&0&0\\0&b&0\\0&0&0\end{bmatrix}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}=\begin{bmatrix}0&0&a\\0&0&0\\0&0&0\end{bmatrix}$$
and similarly we find three other non null matrices with $e_{21},e_{22},e_{23}$. Rest of the matrices will be null matrices after taking product with $e_{31},e_{32},e_{33}$.
Since the range space of $T$ is spanned by $6$ linearly independent matrices so $Rank(T)=6$ and hence $Nullity(T)=9-6=3$.
Method 2: Via Rank-Nullity theorem.
This is much simpler. Let, $T(P)=0$(Null matrix) for some $P=$$\begin{bmatrix}p&q&r\\s&t&u\\v&w&x\end{bmatrix}$ and $p,q...,x\in\Bbb R$. Then, $$\begin{bmatrix}a&0&0\\0&b&0\\0&0&0\end{bmatrix}\begin{bmatrix}p&q&r\\s&t&u\\v&w&x\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
$$\implies\begin{bmatrix}ap&aq&ar\\bs&bt&bu\\0&0&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$
$$\implies p=q=r=s=t=u=0$$ and $v,w,x$ are non zero. So, $Ker(T)=span \left\{ \begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&0\\0&1&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}\right\}$. Now applying Rank-Nullity theorem $dim(M_{3\times 3}(\Bbb R))=dim(Range(T))+dim(Ker(T))\implies 9=Rank(T)+3\implies Rank(T)=6$.
Hope this works.
