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

hw3/hw3.tex

@@ -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}