linear transformation $T_1T_2$ is bijective. Then what can we say about the rank of $T_1$ and $T_2$

For $n\ne m$, let $T_1:\mathbb{R}^n\to \mathbb{R}^m$ and $T_2:\mathbb{R}^m\to \mathbb{R}^n$ be linear transformations such that $T_1T_2$ is bijective. Then what can we say about the rank of $T_1$ and $T_2$

My reasoning: The rank of $T_1$ can at most be $n$, and the rank of $T_2$ can be at most $m$. Now rank of $T_1T_2$ must be at least $n$ otherwise $T_1T_2$ becomes injective. What now?

• is $n<m$? Otherwise, $T_1T_2$ cannot be injective. Jan 28 '16 at 5:32
• Notice that $T_1T_2:\mathbb{R}^m\to\mathbb{R}^m$, so it is injective iff surjective. Jan 28 '16 at 5:33
• Also, $\ker{T_2}\subset\ker{T_1 T_2}$ Can you proof that? Jan 28 '16 at 5:35
• What would be the rank of $T_2$, then? Jan 28 '16 at 5:36
• Sorry. I meant $n>m$. Jan 28 '16 at 5:40

As you've alluded to, the rank of $T_1 T_2$ is $\min\{m,n\}$ in general. If $T_1 T_2$ is bijective, it's rank is exactly $m$. Hence $n \geq m$. You've required $m \neq n$, so $n > m$.

The rank of any linear map is bounded above by the dimensions of both its domain and target space; hence the rank of $T_i$ is $m$ for $i =1,2$.

(Note that $T_1 T_2$ is injective if it's bijective! For linear maps between finite dimensional vector spaces, injectivity, surjectivity, bijectivity, and having a trivial kernel are equivalent.)

• Actually the questions asks to choose from the following 1. rank of T1=n and rank of T2=m. 2. rank of T1=m and rank of T2=n. 3. rank of T1=n and rank of T2=n. 4. rank of T1=m and rank of T2=m.
– Mix
Jan 28 '16 at 5:53
• Then the ranks of each must be $m$. I edited my answer to include this. Jan 28 '16 at 5:58
• But suppose n<m then rank of T1 can be at most n. How can rank of each be m. Am I wrong?
– Mix
Jan 28 '16 at 6:03
• It can't be the case that $n < m$. Do you know the rank-nullity theorem, $\dim \im T + \dim\ker T = \dim V$, where $V$ is the domain of the linear map $T$? Jan 28 '16 at 6:07