# Sufficient condition for a matrix to be PSD

Let us assume that matrix $$S\in \mathbb{R}^{n\times n}$$ is symmetric. We generate a random basis $$B = \{b_1,\dots, b_n\}$$.

We can assume that entries of each vector $$b_i$$ are randomly generated by iid normal distribution $$𝑁(0,\sigma^2)$$. Then we apply the Gram Schmidt process.

Can we say that matrix $$S$$ is PSD with probability one if for all $$i\in [n]$$ we have $$b_i^T S b_i \ge 0$$?

The short answer is no ...

I realized that this is not true by the following example:

Suppose symmetric matrix $$S \in \mathbb{R}^{2\times2}$$ is full rank. Its SVD decomposition is $$S = \lambda_+ v_+ v_+^T + \lambda_- v_- v_-^T$$ such that $$\lambda_-<0<\lambda_+$$ are its eigenvalues and vectors $$v$$'s are the corresponding eigenvectors.

Suppose $$b\in \mathbb{R}^2$$ is a unit one random vector with angle $$\theta_+$$ and $$\theta_-$$ with the corresponding eigenvectors. Then we have

\begin{aligned} b^T S b &= \lambda_+ \cos^2 \theta_+ + \lambda_- \cos^2 \theta_- \\ % &= \lambda_+ \cos^2\left(\pi/2 - \theta_-\right) + \lambda_- \cos^2 \theta_-\\ % &=\lambda_+ \sin^2 \theta_- + \lambda_- \cos^2 \theta_-\\ % &= \lambda_+ + (\lambda_- - \lambda_+) \cos^2 \theta_- < 0. \end{aligned}

We can conclude that $$\theta_- < \arccos\left(\sqrt{\frac{\lambda_+}{\lambda_+ - \lambda_-}}\right) \le \arccos\left(\sqrt{\frac{\kappa}{\kappa + 1}}\right)$$ with $$\kappa$$ being the condition number of $$S$$. So, depending on the condition number, the random vector $$b$$ needs to be tilted towards $$v_-$$. Thus a uniform distribution does not work.

In a higher dimension, we can divide eigenvalues into two sets of positive and negative ones and use the fact that any linear combination of their corresponding eigenvectors is orthogonal.

Now, the question is how to generate vectors $$\pmb{b}$$'s having an upper bound on the ratio of max and min absolute value of non-zero eigenvalues?

• What is a matrix with probability $1$? Sep 21, 2022 at 0:02
• It is not clear what you mean by random, but take $A$ to be [1 -2 ; -2 1]. Then this is positive on the axes, hence for all bases 'nearby', so the almost surely seems unlikely unless you have additional constraints on the distribution. Sep 21, 2022 at 0:10

The question doesn't really make sense because you don't specify a probability distribution over the matrix $$S$$, but in any case, we can say the following: if $$S$$ is has both positive and negative eigenvalues and our random vectors are chosen, say, uniformly on the unit sphere (this is equivalent to using the Gaussian distribution you propose), then with positive probability $$b^T S b \ge 0$$ (e.g. $$b$$ can be any vector in a sufficiently small open neighborhood of an eigenvector of $$S$$ with positive eigenvalue), so with positive probability $$b_i^T S b_i \ge 0$$ for any finite sequence of iid vectors $$b_i$$.