# Which matrices are covariances matrices?

Let $V$ be a matrix.

What conditions should we require so that we can find a random vector $X = (X_1, \dots, X_n)$ so that $V = Var(X)$?

Of course necessary conditions are:

• All the elements on the diagonal should be positive
• The matrix has to be symmetric
• $v_{ij} \le \sqrt{v_{ii}v_{jj}}$ (Because of $Cov(X_i, X_j) \le \sqrt{Var(X_i) Var(X_j)})$

But I am sure these are not sufficient as I have a counterexample.

So what other properties we should require on a matrix so that it can be considered a covariance matrix?

## 1 Answer

I think I cleared this up sufficiently.

Okay, so

1) If $V$ is not semi definite positive, then such a vector $X$ does not exists. (Since all covariances matrix are semi definite positive)

2) If $V$ is symmetric semidefinite positive, then such an $X$ exists! [0]

This implies that

$$\text{exists a random vector X: V = Cov(X)} \iff \text{V is symmetric positive semidefinite}$$

Since we know that those I listed in the question are necessary condition for $V$, we deduce that all symmetric semidefinite positive matrices have elements on the diagonal $\ge 0$ and are such that $v_{ij} \le \sqrt{v_{ii}v_{jj}}$.

These are not sufficient though for a matrix to be semidefinite positive but sufficient conditions are well known, after all.

[0] Proof

Since $V$ is symetric is possible to find an orthogonal matrix $Q$ such that $V = QDQ^T$, where $D$ is a diagonal matrix whose values are the eigenvalues of $V$. If $V$ is semipositive definite the elements of $D$ are all $\ge 0$, hence we can find $X$ such that $D = Cov(X)$ (just take all the variables independent with the specified variance)

It follows that the random vector $QX$ has covariance equal to $$Cov(QX) = QCov(X)Q^T = QDQ^T = V$$

• There's something unclear about your proof: where exactly did you use the condition $v_{ij} \leq \sqrt{v_{ii}v_{jj}}$? Is this a consequence of the positive semidefinite/symmetric matrix condition? Beacuse if not, you there's something wrong in your conclusion. – RandomGuy Oct 31 '16 at 14:32
• @RandomGuy I didn't. I showed that if a matrix is symmetric, positive semidefinite, then it can be written as $Cov(X)$; therefore it must hold that $v_{ij} \le \sqrt{v_{ii}v_{j}}$ for every symmetric semidefinite positive matrix, because it holds for $Cov(X)$. You can consider this a proof of this statement. Of course, if the statement is false, then my proof is invalid :) – Ant Nov 1 '16 at 20:04
• yes, now it makes sense. My question was about the fact that the statement of your original question included that inequality as a condition and in the answer you answer your own question, which seemed to imply that you were using that condition as a hypothesis. – RandomGuy Nov 2 '16 at 8:18