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Author SHA1 Message Date
Claudio Maggioni (maggicl)
d897a331eb hw3: done 1,2,3,4,5a,6 2020-04-23 16:00:22 +02:00
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@ -268,7 +268,7 @@ $$A_{1,1} = A_{1,1}^T$$
and thus $A_{1,1}$ is shown to be symmetric.
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 $p < k \leq n$.
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 $p < k \leq n$.
Then: