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. $AA^{-1}_{right} = I^{n \times n}$?

  • 1
    $\begingroup$ 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 $I_m$. $\endgroup$ – Dirk Mar 28 at 14:25
  • $\begingroup$ @Shuumatsu: Note that $rank(A) \leq \min(m,n)$. $\endgroup$ – Moritz Mar 28 at 14:32
  • $\begingroup$ @Dirk Thanks for your commenting. More details are added to show why I think its bijective. Can you show me where goes wrong? $\endgroup$ – Shuumatsu Mar 28 at 14:47
  • $\begingroup$ My answer give an example where it is not bijective. $\endgroup$ – Lee Mosher Mar 28 at 15:12
  • $\begingroup$ @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? $\endgroup$ – 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 $\begin{pmatrix}1 \\ y \end{pmatrix}$.


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