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Found in: Page 401

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Cholesky factorization for ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrices. Show that any positive definite ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrix A can be written uniquely ${\mathbit{A}}{\mathbf{=}}{\mathbit{L}}{{\mathbit{L}}}^{{\mathbf{T}}}$ as where L is a lower triangular ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrix with positive entries on the diagonal. Hint: Solve the equation $\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]{\mathbf{=}}\left[\begin{array}{cc}x& 0\\ y& z\end{array}\right]\left[\begin{array}{c}\begin{array}{cc}x& y\\ 0& z\end{array}\end{array}\right]$

Every positive definite 2 x 2 matrix

$\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{b}& \mathrm{c}\end{array}\right]$

has the form $A=L{L}^{T}$ in Cholesky factorization, where,

$L=\left[\begin{array}{ccc}\sqrt{a}& 0& \\ b/\sqrt{a}& \sqrt{ac-{b}^{2}}& /\sqrt{a}\end{array}\right]\begin{array}{}\\ \end{array}$

is a positive diagonal matrix with lower triangular entries

See the step by step solution

## Step 1: Given Information:

$\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{b}& \mathrm{c}\end{array}\right]=\left[\begin{array}{cc}\mathrm{x}& 0\\ \mathrm{y}& \mathrm{z}\end{array}\right]\left[\begin{array}{cc}\mathrm{x}& \mathrm{y}\\ 0& \mathrm{z}\end{array}\right]$

## Step 2: Estimating the matrix:

Consider the matrix below:

$\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{b}& \mathrm{c}\end{array}\right]$

be a matrix with a positive definite value. Let

$L=\left[\begin{array}{cc}x& 0\\ y& z\end{array}\right]$

if the lower triangular matrix is 2 x 2, then

$L{L}^{T}=\left[\begin{array}{cc}x& 0\\ y& z\end{array}\right]\left[\begin{array}{cc}x& y\\ 0& z\end{array}\right]=\left[\begin{array}{cc}{x}^{2}& xy\\ yx& {y}^{2}+{z}^{2}\end{array}\right]$

If there is a L with positive diagonal elements, i.e. $x,y,z\in 0$ and $x,z>0$ such that $A=L{L}^{T}$, we say A has Cholesky factorization. If we consider $A=L{L}^{T}$, we get

$\left[\begin{array}{cc}a& b\\ b& c\end{array}\right]=\left[\begin{array}{cc}{x}^{2}& xy\\ yx& {y}^{2}+{z}^{2}\end{array}\right]⇒a={x}^{2},yz=b,y+{z}^{2}=c$

$⇒{\mathrm{x}}^{2}=\mathrm{a},\mathrm{y}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}}},{\mathrm{z}}^{2}=\mathrm{c}-\frac{{\mathrm{b}}^{2}}{\mathrm{a}}=\frac{\mathrm{ac}-{\mathrm{b}}^{2}}{\mathrm{a}}$

Using A's positive definiteness, a>0 and $\mathrm{ac}-{\mathrm{b}}^{2}>0$ , we obtain

role="math" localid="1659671492195" $x=\sqrt{\mathrm{a}}>0,\mathrm{y}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}}}\in \mathrm{ℝ}\mathrm{and}\mathrm{z}=\sqrt{\frac{\mathrm{ac}-{\mathrm{b}}^{2}}{\mathrm{a}}}>0$

For a $2×2$ positive definite matrix, this is the case.

$\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{b}& \mathrm{c}\end{array}\right]$

A lower triangular matrix exists.

$L=\left[\begin{array}{cc}\sqrt{a}& 0\\ b/\sqrt{a}& \sqrt{ac-{b}^{2}}/\sqrt{a}\end{array}\right]$

It has a Cholesky factorization since it has positive diagonal elements and $A=L{L}^{T}$.

## Step 3: Determining the Result:

Every positive definite 2 x 2 matrix

$\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{b}& \mathrm{c}\end{array}\right]$

has the form $A=L{L}^{T}$ in Cholesky factorization, where,

$\mathrm{L}=\left[\begin{array}{cc}\sqrt{\mathrm{a}}& 0\\ \mathrm{b}/\sqrt{\mathrm{a}}& \sqrt{\mathrm{ac}-{\mathrm{b}}^{2}}/\sqrt{\mathrm{a}}\end{array}\right]$

is a positive diagonal matrix with lower triangular entries