# The existence of a matrix?

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!

• You mean $rank(A)$? Dec 4, 2018 at 8:02
• @Yasmin, $A$ needs not be square, I think Dec 4, 2018 at 8:03
• @Yasmin, that's my point, $A$ is $n \times r$ and $n$ and $r$ need not be equal to each other. Dec 4, 2018 at 8:10
• @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 $n \neq r$. Dec 4, 2018 at 10:41

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.

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?

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.

• The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4, 2018 at 8:46
• @A.Γ. Of course. I was editing the answer when you commented it. Dec 4, 2018 at 8:50
• OP specified that $\operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$. Dec 4, 2018 at 17:49
• @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. Dec 4, 2018 at 18:17
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 $$r, $$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.