# What's the meaning of right inverse matrix

For a given matrix $$A \in R^{m \times n}$$, it shows 4 subspace: row space, col space, null space and left null space.

When $$rank(A) = n$$, $$A$$ has a left inverse. When $$rank(A) = m$$, $$A$$ has a right inverse.

$$A$$ can be seen as a function from $$R^n$$ to col space.

$$x$$ and $$y$$ are different vectors in row space and $$x - y$$ should also be in row space. If they are mapped to the same value in col space, then $$x-y$$ is in the null space, which leads to a contradiction, since row space is perpendicular to the null space. So $$A$$ is injective.

When $$rank(A) = n$$, $$R^n$$ equals to row space, so it's a bijective function.

Then the left inverse matrix can be seen as an inverse function which is from col space to $$R^n$$. $$A^{-1}_{left}A = I^{m \times m}$$ means composing a function with its inverse will get the identity function.

But how about the right inverse matrix. $$AA^{-1}_{right} = I^{n \times n}$$?

• Two remarks: First, the function will not be bijective, unless $m = n$. Second, the matrix $I$ is not the same in both cases, one time it is the $n \times n$ identity matrix, one time it is the $m \times m$ one. I would suggest to make this clear, e.g. by calling them $I_n$ and $I_m$. – Dirk Mar 28 at 14:25
• @Shuumatsu: Note that $rank(A) \leq \min(m,n)$. – Moritz Mar 28 at 14:32
• @Dirk Thanks for your commenting. More details are added to show why I think its bijective. Can you show me where goes wrong? – Shuumatsu Mar 28 at 14:47
• My answer give an example where it is not bijective. – Lee Mosher Mar 28 at 15:12
• @LeeMosher sry for the late reply. Because I'm new to linear algebra, I need some time to ponder your explanation. But it seems that your answer doesn't mention the left inverse. If you are saying that "When $rank(A) = m$, $A$ is not bijective" and my proof for "$A$ is bijective when $rank(A) = n$" is true? – Shuumatsu Mar 28 at 15:33

One issue is that "the" right inverse might not be unique. For example, the matrix $$\begin{pmatrix} 1 & 0\end{pmatrix}$$, which defines a linear transformation $$\mathbb R^2 \to \mathbb R^1$$ given by the formula $$f(x,0)=x$$, has many right inverses: each of them has the form $$\begin{pmatrix} 1 \\ y \end{pmatrix}$$.
In the special case where $$n \ge m$$ and $$A$$ is an $$m \times n$$ matrix of rank $$m$$, there is a unified picture for all of the right inverses of $$A$$ which goes like this.
In $$\mathbb R^n$$ consider $$ker(A)$$, the kernel of the matrix $$A$$, consisting of all column vectors $$v \in \mathbb R^n$$ such that $$Av=0$$. Consider an $$n \times m$$ right inverse $$B$$, so $$AB=I$$. Consider $$col(B) \subset \mathbb R^n$$, also known as the image of $$B$$. One can see without too much trouble that $$ker(A)$$ and $$col(B)$$ form complementary subspaces of $$V$$: every vector in $$V$$ can be expressed uniquely as the sum of a vector in $$ker(A)$$ and a vector in $$col(B)$$. The unified picture is that the right inverses of $$A$$ correspond bijectively with the subspaces of $$V$$ that are complementary to $$ker(A)$$.
In the example above, the kernel of the matrix $$A = \begin{pmatrix}1 & 0 \end{pmatrix}$$ is the $$y$$-axis which is the subspace of $$\mathbb R^2$$ spanned by the column vector $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$.
Each of the right inverse matrices of $$A$$, i.e. each of the matrices $$\begin{pmatrix} 1 \\ y \end{pmatrix}$$, corresponds to a complementary subspace of the $$y$$-axis, namely the non-vertical line spanned by the column vector $$\begin{pmatrix}1 \\ y \end{pmatrix}$$.