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) ?
$\endgroup$ 22 Answers
$\begingroup$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
$$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0\\ \end{pmatrix} $$
$$ B = A^T $$
$\endgroup$ 2 $\begingroup$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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