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Let $A$ be a $4 \times 7$ real matrix and $B$ be a $7 \times 4$ real matrix such that $AB=I_4$. Which of the following are true?

1) rank$(A)=4$

2) rank$(B)=7$

3) nullity$(B)=0$

4) $BA=I_7$.

My attempt:

$4=\operatorname{rank}(AB) \leq \min\{{\operatorname{rank}(A),\operatorname{rank}(B)}\}$. So $\operatorname{rank}(A)$ must be $4$. It shows that 1) is true

2) is false by Dimension theorem

How to check 3) and 4) ?

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  • $\begingroup$ Do you know the rank-nullity theorem? Think about the rank of $B$, using the same reasoning you used to answer the first question. This will help you with the third and fourth. $\endgroup$ – kccu May 14 '17 at 2:55
  • $\begingroup$ By the same reasoning you used to prove 1), you can see that 4) is always false (either from 1 or from the falsity of 2). $\endgroup$ – Marc van Leeuwen May 14 '17 at 7:17
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Check (3):

If $nullity(B) \neq \{0\}$ there exists $x \neq 0$ such that $Bx = 0$. Premultiplying this last equality by $A$ you get $0 = ABx = Ix = x$, a contradiction.

Check (4):

Take

$$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0\\ \end{pmatrix} $$

$$ B = A^T $$

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  • $\begingroup$ Very clear! Thank you $\endgroup$ – user444830 May 14 '17 at 3:05
  • $\begingroup$ You're welcome :) $\endgroup$ – rrogerr May 14 '17 at 3:09
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You're correct for the first two. For the third one, your reasoning for the first part also shows that $B$ has rank no smaller than $4$ (and since $B$ has rank no more than $4$, $B$ had rank $4$). Consequently, by the rank-nullity theorem (which tells us that $\mathrm{rank}\, B+\mathrm{null}\, B = 4$), the nullity of $B$ is $0$. For the fourth question, this is impossible, since the rank of $B$ is less than $7$.

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  • $\begingroup$ Oh! Thanks....I got it $\endgroup$ – user444830 May 14 '17 at 3:05

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