I wanted to find real matrices $A$ and $B$ such that:$$(BA)^\dagger\neq A^\dagger B^\dagger$$

whereas $A^\dagger$ denotes the Moore-Penrose pseudoinverse of a matrix.

I tried some things and ended up with:

$$ B = \begin{pmatrix} 0 & 0\\ 1 & 2\end{pmatrix} , A=\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} $$

for these matrices I get $$A^\dagger=\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix}, B^\dagger = \begin{pmatrix} 0 & \frac 15 \\ 0 & \frac 25\end{pmatrix}, A^\dagger B^\dagger = \begin{pmatrix} 0 & \frac 15 \\ 0 & 0\end{pmatrix}$$ but $$(BA)^\dagger=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}$$

so these two are obviously not equal. I found on Wikipedia that the equality stated above holds, when for example $B$ or $A$ has orthonormal columns or rows.

What I am trying to find out here is: Is there a criterion such that the equality does $\bf not$ hold? When trying to find such matrices I just stumbled upon these by accident, say I just picked random matrices which did not fulfill the requirements that the equality automatically holds.

Any help is greatly appreciated!

  • $\begingroup$ Why do you think that there is a general criterion? $\endgroup$ Jun 13, 2017 at 19:40
  • $\begingroup$ @DietrichBurde I don't know, I just thought since there is a criterion wich assures you that the equality holds, maybe there are some type of functions where the Moore Penrose behaves differently in this case... But that was just pure intuition $\endgroup$
    – Jack4t3
    Jun 13, 2017 at 19:47

1 Answer 1


Obviously, you need at least one of your matrices to be not invertible. So let's try it for $2 \times 2$ matrices where $A$ is not invertible but $B$ is. Thus $A$ is of the form $v w^T$ where $v$ and $w$ are column vectors, which I'll suppose are nonzero. It turns out that $$A^\dagger = \frac{w v^T}{ (v^T v) (w^T w)}$$ so that $$ A^\dagger B^\dagger = \frac{w v^T B^{-1}}{(v^T v) (w^T w)} $$ while $B A = (B v) w^T$ so $$ (B A)^\dagger = \frac{w v^T B^T}{((Bv)^T B v) (w^T w)}$$ Thus the columns of both $A^\dagger B^\dagger$ and $(BA)^\dagger$ are scalar multiples of $w$, but the rows are in the one case multiples of $v^T B^{-1} = ((B^{-1})^T v)^T$, and in the other case multiples of $v^T B^T = (B v)^T$. If $(B^{-1})^T$ is not a scalar multiple of $B$, then for almost all $v$ these vectors are not scalar multiples of each other and we will have $A^\dagger B^\dagger \ne (BA)^\dagger$.

  • $\begingroup$ If $(B^{-1})^T = \lambda B$ for some scalar $\lambda$ then $B$ is an orthogonal matrix right? (Orthonormal if $\lambda = 1$). I think the OP mentioned this case. $\endgroup$
    – jnez71
    Jun 13, 2017 at 22:26
  • $\begingroup$ Orthogonal matrix is $(B^{-1})^T = B$. $\endgroup$ Jun 13, 2017 at 23:22

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