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Let $A$ be a symmetric positive definite invertible real valued matrix. Then we can write $A=SS^T$ where $S$ is a lower triangular matrix with positive entries on its diagonal. This decomposition is called the Cholesky decomposition.

How do we know that the entries of $S$ are themselves real-valued? I need this property in order to solve a problem related to the Courant-Fischer min-max theorem.

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  • $\begingroup$ you should add the symmetric condition $\endgroup$ – Ahmad Bazzi Dec 13 '19 at 11:30
  • $\begingroup$ $S$ can be computed by forward substitution. The substitution process does not require field extension. Therefore $S$ is real. $\endgroup$ – user1551 Dec 13 '19 at 13:52
  • $\begingroup$ What is "forward substitution" ? $\endgroup$ – Christian Singer Dec 13 '19 at 14:43
  • $\begingroup$ "Forward substitution" refers to the practice of solving an indexed set of equations iteratively, in which we substitute the solution to the $(k-1)$-th equation into the $k$-th equation and solve the latter. You may think of it as mathematical induction. $\endgroup$ – user1551 Dec 14 '19 at 11:12
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Let $A_k$ be the leading principal $k\times k$ submatrix of $A$. Clearly, $A_1=S_1S_1^T$ where $S_1=\sqrt{A_1}$ is a real $1\times1$ lower triangular matrix. Now suppose that for some $k$, $A_{k-1}=S_{k-1}S_{k-1}^T$ for some real lower triangular matrix $S_{k-1}$. Since $A_{k-1}$ is positive definite, $S_{k-1}$ is nonsingular. Also, as $u^TA_ku>0$ for every nonzero vector $u$, if we write $A_k=\pmatrix{A_{k-1}&v_k\\ v_k^T&a_k}$ and put $u=(-v_k^TA_{k-1}^{-1},\,1)^T$, we obtain $a_k-v_k^TA_{k-1}^{-1}v_k>0$. Therefore the equation $$ \pmatrix{A_{k-1}&v_k\\ v_k^T&a_k} =\pmatrix{S_{k-1}&0\\ x^T&s}\pmatrix{S_{k-1}^T&x\\ 0&s}\tag{1} $$ has a unique solution $x=S_{k-1}^{-1}v_k,\,s=\sqrt{a_k-x^Tx}=\sqrt{a_k-v_k^TA_{k-1}^{-1}v_k}$. This means $A_k=S_kS_k^T$ for some real lower triangular matrix $S_k$. So, if we start from $A_1$ and keep solving $(1)$ for $k=2,3,\ldots$, we see that $A=SS^T$ for some real lower triangular matrix $S$.

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