# How can I find the rank of this linear transformation

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.

• The last part 'so we can simply consider the dimension of all linear transformations from $\Bbb R^3$ to $Q$' is not clear. First, I guess you mean all linear maps to the image of $Q$. Aug 18, 2019 at 18:44

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 $$\operatorname {rank} (\operatorname T) = (9-3) =6$$
Assume w.l.o.g. that $$q_1,q_2$$ are linearly independent columns of $$Q$$ and $$q_3=\alpha q_1+\beta q_2$$.
Then $$PQ=\big[Pq_1\,\mid\, Pq_2 \,\mid\, \alpha Pq_1+\beta Pq_2\big]$$.
Now, for any vectors $$a, b\in\Bbb R^3$$, one can find a matrix $$P$$ such that $$Pq_1=a$$ and $$Pq_2=b$$ (think about e.g. basis extensions and basis transformations).
It means that exactly the $$3\times 3$$ matrices of the form $$\big[a\mid b\mid\alpha a+\beta b\big]$$ will be in the image of the mapping $$P\mapsto PQ$$.
And these matrices depend on $$6$$ scalar values (the coordinates of $$a$$ and $$b$$), so indeed, its rank is $$6$$.