Can one know the rank of a matrix product given the rank of one of them?

Question

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?

Answer 1

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

Answer 2

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 $$ \text{rank}\, Q=\text{rank}\,QT. $$