How to obtain another left inverse? I understand that in principle, it can be that infinitely many left-inverses exist to a matrix $A \in \mathbb{C}^{M \times N}$ if $A$ has $rank = N < M$, i.e. is a tall matrix.
One way to compute a left inverse in this case is via the Moore-Penrose Pseudoinverse, i.e. $A_{left-inverse} = (A^*A)^{-1} A^*$. Alternatively, a SVD also gives a left-inverse.
Currently, I am trying to find a different left-inverse from the following matrix :
$S_s = \begin{bmatrix} s_1^T \\ s_2^T \\ s_3^T \end{bmatrix} = \sqrt{\frac{2}{3}} \cdot \begin{bmatrix} 0 & 1 \\ \frac{\sqrt{3}}{2} & - \frac{1}{2} \\ - \frac{\sqrt{3}}{2} & - \frac{1}{2} \end{bmatrix}$
The pseudo-inverse gives me
$V_s = (S_s^* S_s)^{-1} S_s^* = \sqrt{\frac{2}{3}} \cdot \begin{bmatrix} 0 & 0.866 & - 0.866 \\ 1 & -0.5 & -0.5 \end{bmatrix} = \sqrt{\frac{2}{3}} \cdot \begin{bmatrix} 0 & \frac{\sqrt{3}}{2} & - \frac{\sqrt{3}}{2} \\ 1 & -0.5 & -0.5 \end{bmatrix} = S_s^*$
However, I would like to have another left-inverse that differs from that. How do I do that?
Big thanks in advance!
 A: Suppose we want to characterize the left inverses of $A$, which has SVD $A = U \Sigma V^*$ (with square matrices $U,V$). Let $Z$ be a matrix such that the columns of $Z^*$ span the kernel of $A^*$; such a matrix can be attained, for example, by taking $Z^*$ to be the matrix whose columns are the columns of $U$ corresponding to the zero singular values of $A$ (i.e. the last $M-N$ columns). Let $A^L$ be any single left inverse to $A$ (for example, $A^L = A^+$).
It turns out that a matrix $X$ will satisfy $XA = I$ if and only if we can write $X = A^L + YZ$ for some $M \times (N-M)$ matrix $Y$.
For your particular case, $A = S_s$ has orthonormal columns, which means that its MP pseudoinverse is $A^*$. One possible matrix $Z$ is given by $Z = \pmatrix{1&1&1}$. Thus, every left inverse of $A$ will have the form
$$
X = A^* + \pmatrix{y_1\\y_2}Z = A^* + \pmatrix{y_1&y_1&y_1\\y_2&y_2&y_2}.
$$

It's easy to see that because $ZA = 0$, every matrix of the form $X = YZ + A^L$ will be a left inverse of $A$. To see that every left inverse $X$ has this form, note that $(X - A^L)$ satisfies $(X - A^L)A = 0$. Thus, the kernel of $(X - A^L)$ contains the kernel of $Z$, from which we can deduce that there exists a matrix $Y$ such that $YZ = X - A^L$.
