Definition of the Moore-Penrose pseudoinverse

We know that each matrix $A \in M_{m,n}(\mathbb{R})$ has a unique matrix $B \in M_{n,m}(\mathbb{R})$ that suffices the following four axioms:

  1. $A B A = A$,
  2. $B A B = B $,
  3. $B A$ is symmetric,
  4. $A B$ is symmetric.

This matrix $B$ is called the Moore-Penrose pseudoinverse of A and is often denoted by $A^+$.

The least squares method

For given $A \in M_{m,n}(\mathbb{R})$ and $y \in \mathbb{R}^m$, the least squares method says that $x^* \in \mathbb{R}^n$ is a best approximation for the linear equation system $Ax = y$ if and only if the normal equation $A^t A x^* = A^t y$ holds.


In the literature, one often shows that a best approximation can be given by $x^* = A^+ y$. But I noticed that if we have a matrix $B$ which only satisfies the axioms 1. and 4., then $x^* = By$ is already a best approximation.

Proof: $A^t A x^* = A^t (A B) y \stackrel{4.}{=} A^t (AB)^t y = (ABA)^t y \stackrel{1.}{=} A^t y \tag*{$\blacksquare$}$ .


Do we only require the axioms 1. and 4. to determine the Moore-Penrose pseudoinverse uniquely?

Other problems which I think are similar

We know that in order to define a group we only need to require the existence of a left identity element and for each element in the group the existence of a left inverse element. Then the left identity and the left inverse elements are also right indentity and right inverse elements which are also unique, so this definition is coherent with the standard definition of groups which you find in the literature most of the times.

Maybe the same thing is true for the definition of the Moore-Penrose pseudoinverse now.


1 Answer 1


A counterexample is $$A=\pmatrix {0&0\\1&1\\0&0}$$ $$B=\pmatrix{4&3&0\\-4&-2&0}$$ satisfying $1.$ and $4.$ but neither $2.$ nor $3.$

  • $\begingroup$ Thank you for your response. Could you please explain me how you derived this counterexample? For me, this comes out of nowhere. $\endgroup$
    – Diglett
    Jun 3, 2018 at 11:40
  • $\begingroup$ @Diglett I just found it by generating random matrices with PARI/GP. Unfortunately, I have no idea why it exactly is a counterexample. $\endgroup$
    – Peter
    Jun 3, 2018 at 14:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .