Rank of $ T_1T_2$

Question

For n$\ne $ m let $ T_1 :R^n \to R^m $ and $ T_2:R^m\to R^n $ be linear transformations s.t $ T_1T_2 $ is bijective. Find rank of $ T_1$ and$ T_2$.

I tried by fact that because $ T_1T_2 $ is bijective so rank of $ T_1T_2$ is m. I also found that here n cannot be less than m is that right? So probably their ranks are both m, but I'm not sure.

In general what is the concept to decide rank of product of matrices, if their individual ranks are known?

score 4 · Accepted Answer · answered Jan 25 '15 at 10:50

4

$\newcommand{\R}{\mathbb{R}}$The rank of $T_{1}$ and $T_{2}$ is $m$.

This can be seen by noting that the rank is dimension of the image. So the rank of $T_{1}$ has to be $m$, as its image is $\R^{m}$. The rank of $T_{2}$ cannot be bigger than $m$, and cannot be smaller, otherwise the rank of $T_{1}T_{2}$ would also be smaller than $m$.

answered Jan 25 '15 at 10:50

Andreas Caranti

68,873

Sorry, I was composing left-to-right, as I am used to. Will edit for right-to-left presently. (Done!) – Andreas Caranti Jan 25 '15 at 10:54
Why cant rank of t2 be smaller than m? – Foggy Jan 25 '15 at 11:15
Because the image of $T_{2}$ is generated by the images of the $m$ basis vectors of $R^{m}$. – Andreas Caranti Jan 25 '15 at 11:16

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

The hardest part is to observe that $T_1T_2$ maps $\Bbb R^m\to\Bbb R^m$ (because the rightmost factor acts first). Therefore the rank of $T_1T_2$ (presumed bijective) is $m$. Now using that the rank of a composition of linear maps cannot exceed the rank of either of the two factors, nor can the rank of a linear map exceed the dimension of either of the spaces it maps between (after all one could compose with the identity at either end without changing the map, and the previous case applies), one sees that the rank of $T_1$ or that of $T_2$ cannot be less than $m$ (for the first reason) nor greater than $m$ (for the second reason) so it must be $m$.

Rank of $ T_1T_2$

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