# Expected pseudo-inverse of isotropic random matrix

Suppose I have a random $$m \times m$$ matrix $$R \sim \mu$$ that is possibly singular. Is it true that $$E[R] \propto I$$ implies that there exists a scalar $$r_{\mu, m}$$ such that $$E[R^+] = r_{\mu, m}I$$, where $${}^+$$ denotes the Moore-Penrose pseudo-inverse?

If this is false but there are certain conditions that are needed to make this true, then I'd be interested to know those conditions.

For instance it is true when $$R = X^\top X$$ and $$X$$ is $$n \times m$$ normally distributed (at least I have found references for this when $$n\not\in [m-3, m+1]$$). Not sure if this is the only non-trivial case.

• what does $R\sim \mu$ mean? Jan 28 '21 at 12:12
• The random matrix $R$ has distribution $\mu$. Jan 28 '21 at 16:00
• you write $R^+ = r_{\mu, m}I$ --- don't you want to take the expectation value of $R^+$ ? Jan 28 '21 at 17:58
• yes, that's a typo! thanks Jan 28 '21 at 20:27

## 1 Answer

In general no.

If $$X$$ and $$Y$$ are independent scalar random variables satisfying $$\mathbb E X = \mathbb E Y = 1$$ but $$\mathbb E X^{-1} \neq \mathbb E Y^{-1}$$, then the matrix $$R = \begin{pmatrix} X & 0 \\ 0 & Y \end{pmatrix}$$ is a counterexample.