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Suppose that the matrix $A\in{\mathbb{R}}^{n\times r}$, $\textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $B\in{\mathbb{R}}^{r\times n}$, such that $$AB=I_{n}?$$

What is the requirement for the matrix $A$? Thank you!

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    $\begingroup$ You mean $rank(A)$? $\endgroup$
    – Patricio
    Commented Dec 4, 2018 at 8:02
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    $\begingroup$ @Yasmin, $A$ needs not be square, I think $\endgroup$
    – Patricio
    Commented Dec 4, 2018 at 8:03
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    $\begingroup$ @Yasmin, that's my point, $A$ is $n \times r$ and $n$ and $r$ need not be equal to each other. $\endgroup$
    – Patricio
    Commented Dec 4, 2018 at 8:10
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    $\begingroup$ @Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n \neq r$. $\endgroup$
    – lisyarus
    Commented Dec 4, 2018 at 10:41

4 Answers 4

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From $n=rank (AB)\leq\min\{rank (A),rank (B)\}$ and also $rank(A)\leq\min\{n,r\}$ you can deduce something about the size of $A$ and its rank.

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Consider $$ A=\begin{bmatrix}1\\0\end{bmatrix} $$ then $$ AB=\begin{bmatrix}1\\0\end{bmatrix}\begin{bmatrix}b_1 & b_2\end{bmatrix}=\begin{bmatrix}b_1 & b_2\\0 & 0\end{bmatrix}. $$ Do you see the trouble?

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One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.
Solution(s) exist as soon as $n \le r$ and that $rank(A)=n$.
If $n = r$, $B$ is the inverse of $A$.
If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.
Look at the corresponding Wikipendia entry for example.

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  • $\begingroup$ The Moore-Penrose inverse is just one of many other possibilities for $B$. $\endgroup$
    – A.Γ.
    Commented Dec 4, 2018 at 8:46
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    $\begingroup$ @A.Γ. Of course. I was editing the answer when you commented it. $\endgroup$
    – Damien
    Commented Dec 4, 2018 at 8:50
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    $\begingroup$ OP specified that $\operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$. $\endgroup$ Commented Dec 4, 2018 at 17:49
  • $\begingroup$ @FedericoPoloni The problem can only have a solution if $n \le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion. $\endgroup$
    – Damien
    Commented Dec 4, 2018 at 18:17
  • $\begingroup$ I think OP is asking precisely an exercise about this basic point. $\endgroup$ Commented Dec 4, 2018 at 18:37
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Yet another sufficient way of doing this. Since $rank(A) = r$, we have $r\leq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $r\leq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($\neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.

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