hw3: done 1,2,3,4,5a,6
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hw3/hw3.tex
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@ -255,4 +255,35 @@ A_2 = \begin{bmatrix}
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\[L = \begin{bmatrix}1\\4&2\\8&4&3\\3&8&5&4\\\end{bmatrix}\]
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\section*{Question 5}
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\subsection*{Point a)}
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First of all, to show that $A_{1,1}$ is symmetric, we say:
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\[\begin{bmatrix}A_{1,1} & A_{1,2}\\A_{2,1} & A_{2,2}\\\end{bmatrix} = A = A^T = \begin{bmatrix}A_{1,1}^T & A_{2,1}\\A_{1,2} & A_{2,2}^T\\\end{bmatrix}\]
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Therefore we can say that
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$$A_{1,1} = A_{1,1}^T$$
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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:
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$$ x^TAx = \sum_{j=1}^n x_j \cdot \sum_{i=1}^n x_i a_{i,j} = \sum_{j=1}^p x_j \cdot \sum_{i=1}^n x_i a_{i,j} +
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\sum_{j=p+1}^n 0 \cdot \sum_{i=1}^n 0 a_{i,j} =
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\sum_{j=1}^p x_j \cdot (\sum_{i=i}^p x_i a_{i,j} +
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\sum_{i=p+1}^n 0 a_{i,j}) =$$
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$$=\sum_{j=1}^p x_j \cdot \sum_{i=i}^p x_i a_{i,j} =
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\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}
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A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T$$
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Then:
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$$\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T>0$$
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which is the definition of positive definiteness for $A_{1,1}$.
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\end{document}
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