A real-valued$\Rightarrow$Matrix $S$ of Cholesky-decomposition $A=SS^T$ is real valued.

linear algebramatricesmatrix decompositionpositive definitesymmetric matrices

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.

Best Answer

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 $S$.