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  "body": "\n## Statement\n\nLet $B$ be an $m \\times n$ complex matrix.\nThen the $n \\times n$ matrix $B^* B$ is [[positive semidefinite|01MMVW8CKC]], where $B^*$ is the [[conjugate transpose|0DXKE10PHF]].\nIt is positive definite if and only if the columns of $B$ are linearly independent.\n\n## Remarks\n\nThe proof: $x^* B^* B x = (Bx)^* (Bx) = \\norm{Bx}^2 \\ge 0$.\nFor a real matrix $B$, the same holds for $B^T B$.\nEach [[Gram matrix|00SPJSDT3Z]] has this form, so Gram matrices are positive semidefinite.\n\n## Explanations\n\n### L1 human\n\nMatrices of the form $B^T B$ can never give negative values of $x^T A x$.\nThe value is the squared length of $Bx$.\nIt is zero only when $Bx = 0$.\n",
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