TITLE: Uniqueness of pseudoinverse?
QUESTION [5 upvotes]: Recently, I read a line of reasoning as

Since $A$ (a $3\times 3$ matrix) is of rank $2$, its pseudoinverse is not unique.

May I ask if there is a quickie to show this?

REPLY [5 votes]: As noted by @Algebraic Pavel, the Moore-Penrose pseudoinverse is unique. Perhaps you are thinking of the least squares solution to the problem
$$
 \mathbf{A} x = b.
$$
The column rank defect indicates that the null space $\mathcal{N}(\mathbf{A})$ is not trivial. The least squares minimizers, defined as
$$
  x_{LS} = \text{arg}\!\min\limits_{x\in\mathbb{C}^{n}} \lVert \mathbf{A} x - b \rVert_{2}^{2}
$$
are an affine space given by
$$
 x_{LS} = \mathbf{A}^{\dagger}b + \left(
  \mathbf{I}_{n} - \mathbf{A}^{\dagger} \mathbf{A}
\right) y, \quad y \in\mathbb{C}^{n}.
$$
The arbitrary vector $y$ demonstrates the infinitude of solutions. That is when the matrix rank is less than the number of columns, the least squares solution is not unique.
So in the case of column rank defect the pseudoinverse solution is the point where the affine space of minimizers intersects $\mathcal{N}(\mathbf{A}^{*})$.