# Is covariance matrix positive definite in complex space?

I have a covariance matrix $$\Sigma$$. I know $$\Sigma$$ is positive definite if we are working in real space because for any non-zero $$x \in \mathbb{R}^n$$, $$x^T\Sigma x \gt 0$$ (Edit: I am assuming non-collinearity)

Can I say anything about the sign of $$y^*\Sigma y$$, $$y \in \mathbb{C}^n$$ is a non-zero vector?

## 1 Answer

It probably depends how you define the covariance matrix, and how you define the scalar product. In real space, the transpose and conjugate transpose are the same thing, in complex they are not. Let's say you define the covariance matrix with the latter, and use the standard definition of scalar product in complex space:

$$\boldsymbol{\Sigma}=\mbox{E}\left[\left(\mathbf{X}-\mbox{E}\left[\mathbf{X}\right]\right).\left(\mathbf{X}-\mbox{E}\left[\mathbf{X}\right]\right)^\dagger\right]\quad \boldsymbol{\Sigma}:\,\mathbb{C}^n\to\mathbb{C}^n$$

Where $$\mbox{E}\left[\dots\right]$$ denotes the expectation value. So that ij-th component is:

$$\Sigma_{ij}=\mbox{E}\left[\left(X_i-\mbox{E}\left[X_i\right]\right).\left(X_j-\mbox{E}\left[X_j\right]\right)^{*}\right]$$

So that for any vector $$\mathbf{v} \in \mathbb{C}^n$$:

\begin{align} \mathbf{v}^\dagger.\boldsymbol{\Sigma}.\mathbf{v}=&\mbox{E}\left[\left(\sum_i\left(X_i-\mbox{E}\left[X_i\right]\right)^{*}v_i\right)^{*}\left(\sum_j\left(X_j-\mbox{E}\left[X_j\right]\right)^{*}v_j\right)\right]\\ \ge&0 \end{align}

So positive semi-definite.If the covariance matrix is non-singular, which depends on whether your variables are co-linear, then it will be positive definite