Definition Let $A \in M_{m,n}(\mathbb{R})$ be a matrix. Then a matrix $B \in M_{n,m}(\mathbb{R})$ is called a pseudoinverse of $A$ if we have

  1. $ABA = A$ and
  2. $BAB = B$.

If in addition

  1. $AB$ and $BA$ are symmetric

then we call $B$ a Moore-Penrose pseudoinverse.

In the literature, one shows that every matrix $A \in M_{m,n}(\mathbb{R})$ has a unique Moore-Penrose pseudoinverse $A^+$.

My question: Does the uniqueness still hold if we omit condition 3., i.e. $B$ is a pseudoinverse which is not a Moore-Penrose pseudoinverse?


1 Answer 1


Indeed, if you drop the symmetry condition you might get infinitely many solutions. Take $$ A=\begin{pmatrix} 1& 0\\ 0 &0\end{pmatrix}, \qquad B=\begin{pmatrix} a & b \\ c& d \end{pmatrix}$$ You compute that $ABA=A$ is equivalent to $a=1$ and that $$ ABA=A, \ BAB=B \qquad \text{iff} \qquad bc=d.$$ Hence, there are infinitely many pseudoinverses which are not Moore-Penrose pseudo inverses.


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