TITLE: Can one know the rank of a matrix product given the rank of one of them?
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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
$$
\text{rank}\, Q=\text{rank}\,QT.
$$