91Ó°ÊÓ

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

We've got

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

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,

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

By equating the various components, we arrive at

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

As a result, the Cholesky factorization of A is

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

Cholesky factorization

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