2
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ You mean $rank(A)$? $\endgroup$ – Patricio Dec 4 '18 at 8:02
  • 2
    $\begingroup$ @Yasmin, $A$ needs not be square, I think $\endgroup$ – Patricio Dec 4 '18 at 8:03
  • 1
    $\begingroup$ @Yasmin, that's my point, $A$ is $n \times r$ and $n$ and $r$ need not be equal to each other. $\endgroup$ – Patricio Dec 4 '18 at 8:10
  • 2
    $\begingroup$ @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 Dec 4 '18 at 10:41
6
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The Moore-Penrose inverse is just one of many other possibilities for $B$. $\endgroup$ – A.Γ. Dec 4 '18 at 8:46
  • 1
    $\begingroup$ @A.Γ. Of course. I was editing the answer when you commented it. $\endgroup$ – Damien Dec 4 '18 at 8:50
  • $\begingroup$ OP specified that $\operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$. $\endgroup$ – Federico Poloni Dec 4 '18 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 Dec 4 '18 at 18:17
  • $\begingroup$ I think OP is asking precisely an exercise about this basic point. $\endgroup$ – Federico Poloni Dec 4 '18 at 18:37
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.