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Suppose we have some matrix $Q\in\mathbb{R}^{m\times n}$. We have another matrix $T\in\mathbb{R}^{n\times n}$, which happens to be full-rank.

I'm wondering if it is possible to know the rank of the matrix built as $$\hat{Q}=QT$$

My intuition says that, due to $T$ being an injection, $\mathrm{rank}(\hat{Q})=\mathrm{rank}(Q)$, but I'm not sure about this. I find it easy to arrive to this conclusion if the matrix was $\hat{Q}=TQ$, with $Q\in\mathbb{R}^{n\times m}$ but not with the $T$ at the right.

Any ideas?

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2 Answers 2

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Idea: combine $$ \text{rank}\,\hat Q=\text{rank}\,QT\le\text{rank}\,Q $$ and $$ \text{rank}\,Q=\text{rank}\,\hat QT^{-1}\le\text{rank}\,\hat Q. $$

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The rank is the dimension of the image. Since $T$ is injective and injective maps preserve dimension (as they map linearly independent sets to linearly independent sets), it follows that $T$ is also surjective. So the image of $T$ is $\mathbb R^n$. It follows that the image of $QT$ is equal to the image of $Q$, so in particular $$ \text{rank}\, Q=\text{rank}\,QT. $$

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  • $\begingroup$ No quite. If T maps all vectors to the null space of Q, then the rank of (QT) would be 0. So, it is more complicated. $\endgroup$ May 13, 2018 at 17:09
  • $\begingroup$ Not sure what you mean. $T$ is a bijective map $\mathbb R^n\to\mathbb R^n$; it has no "room" to map all vectors to the kernel of $Q$, as $Q$ maps $\mathbb R^n\to\mathbb R^m$. With $T$ bijective, every possible vector in the domain of $Q$ is offered to $Q$. $\endgroup$ May 13, 2018 at 17:11
  • $\begingroup$ @ulaff.net If $T$ maps all vectors to the null space of $Q$, then the null space of $Q$ would have to be $\mathbb{R}^n$. If this is true, then $Q$ should have rank $0$, thus $Q = \mathbf{0}$, and this is not a pretty interesting case. $\endgroup$
    – Tendero
    May 13, 2018 at 17:13
  • $\begingroup$ You are right. I did not read the question carefully! $\endgroup$ May 14, 2018 at 1:58
  • $\begingroup$ No worries. Happened twice to me today. $\endgroup$ May 14, 2018 at 2:00

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