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  "body": "\n## Proof\n\n1. Let $B$ be an $m \\times n$ complex matrix, and let $A = B^* B$, where $^*$ is the [[conjugate transpose|0EX045E2BP]]. Then $A$ is $n \\times n$.\n2. For matrices $X$ and $Y$ with $XY$ defined, $(XY)^* = Y^* X^*$. This is the [[transpose of a product|09GK5PS824]] with all entries conjugated.\n3. By step 2, $A^* = B^* (B^*)^* = B^* B = A$. So $A$ is [[Hermitian|0EKZCX4K3P]].\n4. For a vector $y \\in \\C^m$, we have $y^* y = \\sum_{i=1}^m \\abs{y_i}^2 \\ge 0$, with equality only if $y = 0$.\n5. Let $x \\in \\C^n$. By step 2, $x^* A x = x^* B^* B x = (Bx)^* (Bx)$.\n6. By steps 4 and 5 with $y = Bx$, we get $x^* A x \\ge 0$. So $A$ is [[positive semidefinite|0880F9EQCP]].\n7. Let $b_1, \\dots, b_n$ be the columns of $B$. Then $Bx = x_1 b_1 + \\cdots + x_n b_n$.\n8. By steps 4 and 5, $x^* A x = 0$ holds if and only if $Bx = 0$.\n9. Suppose the columns are [[linearly independent|0EVHJFTG9Z]], and let $x \\ne 0$. By step 7, $Bx \\ne 0$. By step 8, $x^* A x \u003e 0$. So $A$ is [[positive definite|0QQCG3JWJ4]].\n10. Conversely, suppose the columns are linearly dependent. Then there is $x \\ne 0$ with $x_1 b_1 + \\cdots + x_n b_n = 0$, that is, $Bx = 0$.\n11. By step 8, $x^* A x = 0$ for this $x \\ne 0$. So $A$ is not positive definite.\n12. By steps 9 to 11, $B^* B$ is positive definite if and only if the columns of $B$ are linearly independent.\n\n## Idea\n\nThe number $x^* B^* B x$ is the squared length of the vector $Bx$. A squared length is never negative, and it is zero only for the zero vector. The vector $Bx$ is a combination of the columns of $B$, so it vanishes for some $x \\ne 0$ exactly when the columns are dependent.\n",
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