Suppose $Q \in M_{3 \times 3}\mathbb(R)$ is a matrix of rank $2$.
Let $T : M_{3 \times 3}\mathbb(R) \to M_{3 \times 3}\mathbb(R)$ be the linear transformation defined by $T(P) = PQ$. Then rank of T is:
I have searched the site and found that Find Rank of given Linear transformation. is very similar to my question.
The only difference is that role of matrices are reversed.
But i cannot see how to correctly use the answer given in the original question to solve mine.
Here is what I came up with
since Rank($AB$) $\le$ min(Rank$A$, Rank$B$)
so Rank T = Rank($PQ$) $\le$ min(RankP,Rank $Q$) < Rank(Q)
so we can simply consider the dimension of all linear transformations from $R^3$ to Q
that will give us the rank of T.
hence rank of T = 6
Is this correct ? if not can anyone give me the correct solution please.
Best Answer
Actually, as your linear transformation is $\operatorname T(\operatorname P) = \operatorname{PQ}$ where $\operatorname Q$ is a matrix of $\operatorname {rank} 2$, and hence if you do consider the standard basis of $\operatorname M_3(\mathbb R)$ as $\beta = \{ \operatorname E_{ij} : i,j \in \{1,2,3\} \}$ where $\operatorname E_{11} = $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ \ etc ....$, then you can easily check that $$\operatorname T(\operatorname E_{ij}) = \sum_{k=1}^3 q_{jk}\operatorname E_{ik}$$ for all $i \in \{1,2,3\}$ where $\operatorname Q = (q_{ij})_{3×3} .$ Hence, it turns out that the matrix of transformation of $[\operatorname T]_{st}^{st}$ is a $9×9$ matrix and as $\operatorname {rank} (\operatorname Q) =2$ , hence writing the $9×9$ matrix, there are exactly $3$ linearly dependent columns, hence $\operatorname {rank} (\operatorname T) = (9-3) =6$