# Finding all left inverse of a matrix B

Let $$\mathbf{B} = \left[\begin{array}{cc} 0 & 1 \\ 1 & -3 \\ 0 & 0 \end{array}\right].$$

My task at hand is to all left-inverse matrices of $$\mathbf{B}$$, without using the Moore-Penrose pseudoinverse.

My attempt so far is the following:
I would like to solve the equation $$\mathbf{X}\mathbf{B}=\mathbf{I}$$ i.e $$\left[\begin{array}{lll} x_1 & x_2 & x_3 \\ x_4 & x_5 & x_6 \end{array}\right]\left[\begin{array}{cc} 0 & 1 \\ 1 & -3 \\ 0 & 0 \end{array}\right] =\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$ I do this by solving the system

$$[\mathbf{B}^{\top}|\mathbf{I}] = \left[\begin{array}{ccc|cc} 0 & 1 & 0 & 1 & 0 \\ 1 & -3 & 0 & 0 & 1 \end{array}\right]$$

By doing rowoperations I get the following RREF $$\left[\begin{array}{lll|ll} 1 & 0 & 0 & 3 & 1 \\ 0 & 1 & 0 & 1 & 0 \end{array}\right]$$

Hence $$x_1 = 3, x_2=1, x_3 = s, x_4=1, x_5=0, x_6=t$$ where $$s$$ and $$t$$ are free variables i.e there are infinitely many solutions and all left-inverses can be described by the matrix

$$\left[\begin{array}{lll} 3 & 1 & s \\ 1 & 0 & t \end{array}\right]$$

Is this correct?

• I would say yes May 16, 2021 at 21:24

## 1 Answer

Yes. It is totally correct.

A non-square matrix has always a side-inverse (in this case a left-inverse) if it has full rank. However, the side-inverse does not have to be unique.

Actually I think that a side-invertible non-square matrix has always several side-inverses (maybe an infinite number). The proof of this can be an interesting exercise to do.

Edit: my intuition was wrong. Thanks to the comment below, we know that a matrix has several left inverses if it has more rows than columns, and it has zero left inverses if it has more columns than rows.

• The matrix has several left inverses if it has more rows than columns, and it has zero left inverses if it has more columns than rows.
– 5xum
May 17, 2021 at 9:27
• Great comment! Thank you for the information. May 17, 2021 at 9:38