hw3: done 1,2,3,4,5a,6

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Claudio Maggioni (maggicl) 2020-04-23 15:58:53 +02:00
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@ -255,4 +255,35 @@ A_2 = \begin{bmatrix}
\[L = \begin{bmatrix}1\\4&2\\8&4&3\\3&8&5&4\\\end{bmatrix}\]
\section*{Question 5}
\subsection*{Point a)}
First of all, to show that $A_{1,1}$ is symmetric, we say:
\[\begin{bmatrix}A_{1,1} & A_{1,2}\\A_{2,1} & A_{2,2}\\\end{bmatrix} = A = A^T = \begin{bmatrix}A_{1,1}^T & A_{2,1}\\A_{1,2} & A_{2,2}^T\\\end{bmatrix}\]
Therefore we can say that
$$A_{1,1} = A_{1,1}^T$$
and thus $A_{1,1}$ is shown to be symmetric.
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:
$$ x^TAx = \sum_{j=1}^n x_j \cdot \sum_{i=1}^n x_i a_{i,j} = \sum_{j=1}^p x_j \cdot \sum_{i=1}^n x_i a_{i,j} +
\sum_{j=p+1}^n 0 \cdot \sum_{i=1}^n 0 a_{i,j} =
\sum_{j=1}^p x_j \cdot (\sum_{i=i}^p x_i a_{i,j} +
\sum_{i=p+1}^n 0 a_{i,j}) =$$
$$=\sum_{j=1}^p x_j \cdot \sum_{i=i}^p x_i a_{i,j} =
\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}
A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T$$
Then:
$$\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T>0$$
which is the definition of positive definiteness for $A_{1,1}$.
\end{document}