# When is $(BA)^\dagger\neq A^\dagger B^\dagger$ for the Moore Penrose inverse?

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!

• Why do you think that there is a general criterion? Jun 13, 2017 at 19:40
• @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 Jun 13, 2017 at 19:47

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 $A^\dagger B^\dagger \ne (BA)^\dagger$.
• 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. Jun 13, 2017 at 22:26
• Orthogonal matrix is $(B^{-1})^T = B$. Jun 13, 2017 at 23:22