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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.

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  • $\begingroup$ 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 $Q$. $\endgroup$ – Berci Aug 18 '19 at 18:44
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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 $\operatorname {rank} (\operatorname T) = (9-3) =6$

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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 $6$.

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