Gram matrices are positive semidefinite

Statement

Let $B$ be an $m \times n$ complex matrix. Then the $n \times n$ matrix $B^* B$ is positive semidefinite, where $B^*$ is the conjugate transpose. It is positive definite if and only if the columns of $B$ are linearly independent.

Remarks

The proof: $x^* B^* B x = (Bx)^* (Bx) = \norm{Bx}^2 \ge 0$. For a real matrix $B$, the same holds for $B^T B$. Each Gram matrix has this form, so Gram matrices are positive semidefinite.

Explanations

L1 human

Matrices of the form $B^T B$ can never give negative values of $x^T A x$. The value is the squared length of $Bx$. It is zero only when $Bx = 0$.