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Expert-verified Found in: Page 401 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find the Cholesky factorization of the matrix ${\mathbit{A}}{\mathbf{=}}\left[\begin{array}{ccc}4& -4& 8\\ -4& 13& 1\\ 8& 1& 26\end{array}\right]$

Cholesky factorization

$A=\left[\begin{array}{ccc}2& 0& 0\\ -2& 3& 0\\ 4& 3& 1\end{array}\right]\left[\begin{array}{ccc}2& -2& 4\\ 0& 3& 3\\ 0& 0& 1\end{array}\right]$

See the step by step solution

## Step 1: Given Information:

$R=\left[\begin{array}{ccc}4& -4& 8\\ -4& 13& 1\\ 8& 1& 26\end{array}\right]$

## Step 2: Determining Cholesky factorization of the matrix:

We've got

$\left|{A}^{\left(1\right)}\right|=4>0,\left|{A}^{\left(2\right)}\right|=4\left(13\right)-16=36>0\phantom{\rule{0ex}{0ex}}\left|{A}^{\left(3\right)}\right|=4\left(337\right)+4\left(-112\right)+8\left(-108\right)=36>0$

All of the determinants of main submatrices are positive; hence A is positive definite and has a Cholesky factorization, according to theorem 8.2.5. Let

$\mathrm{L}=\left[\begin{array}{ccc}\mathrm{a}& 0& 0\\ \mathrm{b}& \mathrm{d}& 0\\ \mathrm{c}& \mathrm{e}& \mathrm{f}\end{array}\right]$

be the lower triangular matrix with positive diagonal entries, i.e. a, d, f>0, with $\mathrm{A}={\mathrm{LL}}^{\mathrm{T}}$. Thus,

${\mathrm{LL}}^{\mathrm{T}}=\left[\begin{array}{ccc}\mathrm{a}& 0& 0\\ \mathrm{b}& \mathrm{d}& 0\\ \mathrm{c}& \mathrm{e}& \mathrm{f}\end{array}\right]\left[\begin{array}{ccc}\mathrm{a}& \mathrm{b}& \mathrm{c}\\ 0& \mathrm{d}& \mathrm{e}\\ 0& 0& \mathrm{f}\end{array}\right]=\left[\begin{array}{ccc}{\mathrm{a}}^{2}& \mathrm{ab}& \mathrm{ac}\\ \mathrm{ba}& {\mathrm{b}}^{2}+{\mathrm{d}}^{2}& \mathrm{bc}+\mathrm{de}\\ \mathrm{ca}& \mathrm{cb}+\mathrm{ed}& {\mathrm{c}}^{2}+{\mathrm{e}}^{2}+{\mathrm{f}}^{2}\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{A}={\mathrm{LL}}^{\mathrm{T}}⇒\left[\begin{array}{ccc}{\mathrm{a}}^{2}& \mathrm{ab}& \mathrm{ac}\\ \mathrm{ba}& {\mathrm{b}}^{2}+{\mathrm{d}}^{2}& \mathrm{bc}+\mathrm{de}\\ \mathrm{ca}& \mathrm{cb}+\mathrm{ed}& {\mathrm{c}}^{2}+{\mathrm{e}}^{2}+{\mathrm{f}}^{2}\end{array}\right]=\left[\begin{array}{ccc}4& -4& 8\\ -4& 13& 1\\ 8& 1& 26\end{array}\right]$

By equating the various components, we arrive at

${\mathrm{a}}^{2}=4⇒\mathrm{a}=2\left(>0\right)\phantom{\rule{0ex}{0ex}}\mathrm{ab}=-4⇒\mathrm{b}=-2\phantom{\rule{0ex}{0ex}}\mathrm{ac}=8⇒\mathrm{c}=4\phantom{\rule{0ex}{0ex}}{\mathrm{b}}^{2}+{\mathrm{d}}^{2}=13⇒{\mathrm{d}}^{2}=13-4=9⇒\mathrm{d}=3\left(>0\right)\phantom{\rule{0ex}{0ex}}\mathrm{bc}+\mathrm{de}=1⇒3\mathrm{e}=9⇒\mathrm{e}=3\phantom{\rule{0ex}{0ex}}{\mathrm{c}}^{2}+{\mathrm{e}}^{2}+{\mathrm{f}}^{2}=26⇒{\mathrm{f}}^{2}=26-16=9=1⇒\mathrm{f}=1\left(>0\right)$

As a result, the Cholesky factorization of A is

$\mathrm{A}=\left[\begin{array}{ccc}4& -4& 8\\ -4& 13& 1\\ 8& 1& 26\end{array}\right]=\left[\begin{array}{ccc}2& 0& 0\\ -2& 3& 0\\ 4& 3& 1\end{array}\right]\left[\begin{array}{ccc}2& -2& 4\\ 0& 3& 3\\ 0& 0& 1\end{array}\right]={\mathrm{LL}}^{\mathrm{T}}$

## Step 3: Determining the Result:

Cholesky factorization

$A=\left[\begin{array}{ccc}2& 0& 0\\ -2& 3& 0\\ 4& 3& 1\end{array}\right]\left[\begin{array}{ccc}2& -2& 4\\ 0& 3& 3\\ 0& 0& 1\end{array}\right]$ ### Want to see more solutions like these? 