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@ -268,7 +268,7 @@ $$A_{1,1} = A_{1,1}^T$$
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and thus $A_{1,1}$ is shown to be symmetric.
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Then, since $A$ is positive definite, $x^TAx > 0$ for any vector $x \in R^n$. We choose a vector $x$ such that $x_k = 0$ for any $p < k \leq n$.
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Then, since $A$ is positive definite, $ > 0$ for any vector $x \in R^n$. We choose a vector $x$ such that $x_k = 0$ for any $p < k \leq n$.
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Then:
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