Assume a positive semi-definite $M\times M$ matrix $A$, not with full rank, and an $M\times N$ matrix $X$, where $M>N$. The elements of $X$ are independent, zero-mean complex Gaussian with variance $1/M$.

My question is simple, what is the distribution of $X^H AX$?

From what I have seen, a matrix of form $X^H X$ is Wishart if the rows of $X$ are correlated, but in my case it is the columns.


1 Answer 1


Necessary and sufficient conditions on $A$ for $Q=X^H AX$ to have a Wishart distribution are derived in Wishart and chi-square distributions associated with matrix quadratic forms (1997) and Wishart distributions associated with matrix quadratic forms (2003).

In general, the matrix $Q$ can be decomposed as a linear combination of independent Wishart matrices, see On the distribution of matrix quadratic forms (2012).

  • $\begingroup$ Understood, and many thanks. It was easily verified that for my case it is not Wishart. I have a follow-up question, perhaps a more difficult one. From the last paper you mentioned, we know that $Q=\sum_{k=1}^K \lambda_k W_k$, where $W_k$ are drawn iid from a certain Wishart distribution. My problem is now that I would like to pass $K\to \infty$. My $\lambda_k$ are then defined from a function $\lambda(x)$, i.e., for a given K, we have $Q=(1/K)\sum_{k=1}^K \lambda (k/K) W_k$. Now, what is the limiting distribution of $Q$? $\endgroup$
    – Nicki
    Jul 8, 2017 at 21:12
  • 1
    $\begingroup$ in the limit $K\rightarrow\infty$ the law of large numbers will ensure that $P(Q)$ tends to a delta function at $Q_{\infty}=\mathbb{1}\,M^{-1}\,{\rm tr}\, A$. $\endgroup$ Jul 9, 2017 at 11:28

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