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  "body": "\n## Proof\n\nWe prove existence by induction on $n$, then uniqueness, then the real case. Here $^*$ is the [[conjugate transpose|0EX045E2BP]].\n\n1. For matrices $X$ and $Y$ with $XY$ defined, $(XY)^* = Y^* X^*$. This is the [[transpose of a product|09GK5PS824]] with all entries conjugated.\n2. For $n = 1$, $A = (a)$ and $a = e_1^* A e_1 \u003e 0$ by [[positive definiteness|0QQCG3JWJ4]]. By the [[existence of square roots|068BZBXE0C]], $L = (\\sqrt a)$ works.\n3. Let $n \\ge 2$ and assume existence for size $n - 1$. Since $A$ is [[Hermitian|0EKZCX4K3P]], $A = \\begin{pmatrix} A' \u0026 b \\\\ b^* \u0026 c \\end{pmatrix}$ with $A'$ of size $n - 1$, $b \\in \\C^{n-1}$ and $c \\in \\C$.\n4. The block $A'$ is Hermitian. For $z \\in \\C^{n-1}$ with $z \\ne 0$, let $x = (z, 0)$. Then $z^* A' z = x^* A x \u003e 0$. So $A'$ is positive definite.\n5. By the induction hypothesis, $A' = L' L'^*$ with $L'$ [[lower triangular|0EZSD6T9E0]] and with positive diagonal entries.\n6. By the [[determinant of a triangular matrix|0K1KKMFHYH]], $\\det L'$ is the product of these entries, so it is nonzero. So $L'$ is [[invertible|FBSAPTP1PR]], and so is $L'^*$.\n7. Let $m = L'^{-1} b$ and $w = -(L'^*)^{-1} m$. Let $x \\in \\C^n$ be $w$ followed by the entry $1$, so $x \\ne 0$.\n8. We have $A' w = L' L'^* w = -L' m = -b$. Expanding in blocks gives $x^* A x = w^* A' w + w^* b + b^* w + c = b^* w + c$.\n9. By step 1, $b^* = (L' m)^* = m^* L'^*$. So $b^* w = -m^* L'^* (L'^*)^{-1} m = -m^* m$.\n10. By steps 7 to 9, $s = c - m^* m$ equals $x^* A x$. So $s$ is a positive real number.\n11. Let $d = \\sqrt s \u003e 0$, and let $L = \\begin{pmatrix} L' \u0026 0 \\\\ m^* \u0026 d \\end{pmatrix}$. It is lower triangular with positive diagonal entries.\n12. Block multiplication gives $L L^* = \\begin{pmatrix} L' L'^* \u0026 L' m \\\\ m^* L'^* \u0026 m^* m + d^2 \\end{pmatrix}$.\n13. By steps 5, 7, 9 and 10, this equals $\\begin{pmatrix} A' \u0026 b \\\\ b^* \u0026 c \\end{pmatrix} = A$. This completes the induction.\n14. For uniqueness we need a lemma. Let $M$ be an invertible lower triangular matrix with entries $m_{ij}$. As in step 6, $m_{ii} \\ne 0$ for all $i$.\n15. Fix $j$ and let $y$ be column $j$ of $M^{-1}$, so $My = e_j$. For $i \u003c j$, row $i$ reads $\\sum_{k \\le i} m_{ik} y_k = 0$.\n16. For $i = 1, 2, \\dots, j - 1$ in turn, this gives $m_{ii} y_i = -\\sum_{k \u003c i} m_{ik} y_k = 0$. So $y_i = 0$ for $i \u003c j$, and row $j$ gives $y_j = 1 / m_{jj}$.\n17. So $M^{-1}$ is lower triangular with diagonal entries $1/m_{jj}$.\n18. If $S$ and $T$ are lower triangular, then $(ST)_{ij} = \\sum_k s_{ik} t_{kj}$ has nonzero terms only for $j \\le k \\le i$. So $ST$ is lower triangular with diagonal entries $s_{ii} t_{ii}$.\n19. The same argument shows that a product of [[upper triangular|0MKABWDRNC]] matrices is upper triangular. The conjugate transpose of a lower triangular matrix is upper triangular.\n20. Now let $A = L L^* = M M^*$ with $L$ and $M$ lower triangular with positive diagonal entries. Both are invertible, as in step 6.\n21. Let $D = M^{-1} L$. By steps 17 and 18, $D$ is lower triangular with positive diagonal entries $l_{ii} / m_{ii}$.\n22. From $L L^* = M M^*$ we get $D L^* = M^*$, so $D = M^* (L^*)^{-1}$.\n23. By step 1, $(L^{-1})^* L^* = (L L^{-1})^* = I$, so $(L^*)^{-1} = (L^{-1})^*$. By steps 17 and 19, it is upper triangular.\n24. By steps 19, 22 and 23, $D$ is upper triangular. It is also lower triangular, so $D$ is diagonal with positive diagonal entries $d_{ii}$.\n25. Then $L = MD$, so $M M^* = L L^* = M D D^* M^*$. Multiplying by $M^{-1}$ on the left and $(M^*)^{-1}$ on the right gives $D D^* = I$.\n26. So $d_{ii}^2 = 1$, and $d_{ii} = 1$ since $d_{ii} \u003e 0$. Hence $D = I$ and $L = M$.\n27. Real case: let $A$ be real, and let $\\bar L$ be the matrix of complex conjugates of the entries of $L$.\n28. Then $\\bar L$ is lower triangular with the same positive diagonal entries, and $\\bar L\\, \\bar L^* = \\overline{L L^*} = \\bar A = A$.\n29. By the uniqueness of steps 20 to 26, $\\bar L = L$. So all entries of $L$ are real.\n\n## Idea\n\nSplit off the last row and column. The top left block has a Cholesky factor by induction, and the last row of $L$ is then forced by two equations. The only delicate point is that $c - m^* m$ is positive, and it is the value $x^* A x$ at a well chosen vector $x$.\n",
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