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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?

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

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  • $\begingroup$ 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. $\endgroup$
    – Mix
    Jan 28 '16 at 5:53
  • $\begingroup$ Then the ranks of each must be $m$. I edited my answer to include this. $\endgroup$ Jan 28 '16 at 5:58
  • $\begingroup$ But suppose n<m then rank of T1 can be at most n. How can rank of each be m. Am I wrong? $\endgroup$
    – Mix
    Jan 28 '16 at 6:03
  • $\begingroup$ 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 $T$? $\endgroup$ Jan 28 '16 at 6:07

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