# 数学代写|数值分析代写numerical analysis代考|Variants of LU Factorization

## 数学代写|数值分析代写numerical analysis代考|Positive-Definite Matrices

Positive-definite matrices arise in a number of contexts: in statistics where, for example, variance-covariance matrices are positive definite; in optimization where convex functions can be identified by having positive-definite Hessian matrices; in physics, quadratic energy functions are generated by positive-definite matrices; in partial differential equations, where the discretization of elliptic equations leads to positivedefinite matrices.

We take (2.1.9) as the definition that $A$ is positive definite. Many authors assume that when a matrix is described as positive definite, it must also be symmetric. Here we do not. ${ }^1$ If a matrix is both positive definite and symmetric, we will say so explicitly. We do assume, unless otherwise stated, that a matrix is real. For a complex matrix $A$ we modify definition (2.1.9) to
(2.1.10) $\operatorname{Re} \bar{z}^T A z>0 \quad$ for all complex $z \neq \mathbf{0}$.
Here $\bar{z}$ is the vector $z$ with the entries of $z$ replaced by their complex conjugates. Note that the condition ” $\bar{z}^T A z>0$ ” implies that $\bar{z}^T A z$ is real. In the complex case, instead of asking for $A$ to be symmetric $\left(A^T=A\right)$, we ask for $A$ to be Hermitian:
(2.1.11) $\bar{A}^T=A ; \quad$ that is, $\overline{a_{\ell k}}=a_{k \ell} \quad$ for all $k, \ell$.
A real matrix $A$ is positive definite if and only if its symmetric part $\frac{1}{2}\left(A+A^T\right)$ is positive definite. For complex matrices, $A$ is positive definite if and only if $\frac{1}{2}(A+$ $\bar{A}^T$ ). As an example of a positive-definite matrix that is not symmetric, consider
$$A=\left[\begin{array}{ll} 2 & 1 \ 0 & 1 \end{array}\right] .$$
From definition (2.1.9), positive-definite matrices have a number of important properties: positive-definite matrices are invertible; if $A$ and $B$ are positive-definite matrices, so are $A+B, \alpha A$ for $\alpha>0$, and $A^{-1}$. Symmetric positive-definite matrices can also be understood in terms of eigenvalues: all the eigenvalues of a symmetric matrix are real, but the eigenvalues of a symmetric positive-definite matrix are positive: if $A \boldsymbol{v}=\lambda \boldsymbol{v}$ and $\boldsymbol{v} \neq \boldsymbol{0}$ then $\boldsymbol{v}^T A v=\lambda \boldsymbol{v}^T \boldsymbol{v}>0$ so $\lambda>0$. In fact, a symmetric matrix is positive definite if and only if all its eigenvalues are positive.

## 数学代写|数值分析代写numerical analysis代考|LDLT Factorization and BK Factorization

Both $L D L^T$ and BK (Bunch-Kaufman) [39] factorizations involve having a diagonal or block-diagonal matrix $D$, and $L$ matrices that have one’s on the diagonal. The $L D L^T$ factorization of a symmetric positive-definite matrix is equivalent to the Cholesky factorization, as then the diagonal matrix $D$ must have positive diagonal entries, and
\begin{aligned} A & =L D L^T=\left(L D^{1 / 2}\right)\left(L D^{1 / 2}\right)^T, \ \left(D^{1 / 2}\right){k k} & =\sqrt{d{k k}} \text { and }\left(D^{1 / 2}\right)_{k \ell}=0 \text { if } k \neq \ell . \end{aligned}
If $A$ is positive definite, the advantage of the $L D L^T$ factorization is avoiding computing square roots. Square roots take roughly $10-20$ times as long to compute as addition, subtraction, or multiplication in modern architectures, so this could improve performance. On the other hand, there are only $n$ square root computations in com-puting the Cholesky factorization of an $n \times n$ matrix compared with $\sim \frac{1}{3} n^3$ other floating point operations. The cost of these $n$ square roots is small compared to the other floating point operations in Cholesky factorization for $n>10$.

To see how the $L D L^T$ factorization works, consider the recursive decomposition
$$A=\left[\frac{\alpha \mid \boldsymbol{a}^T}{\boldsymbol{a} \mid \widetilde{A}}\right]=\left[\frac{1 \mid}{\ell \mid \widetilde{L}}\right]\left[\begin{array}{cc} \delta & \ & \widetilde{D} \end{array}\right]\left[\frac{1 \mid \ell^T}{\mid \widetilde{L}^T}\right]=L D L^T .$$
From this, we have the equations
\begin{aligned} \delta & =\alpha, \ \delta \boldsymbol{\ell} & =\boldsymbol{a}, \quad \text { so } \boldsymbol{\ell}=\boldsymbol{a} / \delta=\boldsymbol{a} / \alpha \ \tilde{A}-\alpha \ell \ell^T & =\widetilde{L} \widetilde{D} \widetilde{L}^T \quad \text { (recursive } L D L^T \text { factorization). } \end{aligned}

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|Positive-Definite Matrices

(2.1.9) 修改为
(2.1.10) $\operatorname{Re} \bar{z}^T A z>0 \quad$ 对于所有复杂的 $z \neq \mathbf{0}$.

Hermitian:
$(2.1 .11) \bar{A}^T=A$; 那是， $\overline{a_{\ell k}}=a_{k \ell} \quad$ 对全部 $k, \ell$.

$$A=\left[\begin{array}{llll} 2 & 1 & 0 & 1 \end{array}\right]$$

$A \boldsymbol{v}=\lambda \boldsymbol{v}$ 和 $\boldsymbol{v} \neq \boldsymbol{0}$ 然后 $\boldsymbol{v}^T A v=\lambda \boldsymbol{v}^T \boldsymbol{v}>0$ 所以 $\lambda>0$. 事实上，一对称矩阵是正定 的当且仅当它的所有特征值都是正的。

## 数学代写|数值分析代写numerical analysis代考|LDLT Factorization and BK Factorization

$$A=L D L^T=\left(L D^{1 / 2}\right)\left(L D^{1 / 2}\right)^T,\left(D^{1 / 2}\right) k k \quad=\sqrt{d k k} \text { and }\left(D^{1 / 2}\right)_{k \ell}=0 \text { if } k$$

$$A=\left[\frac{\alpha \mid \boldsymbol{a}^T}{\boldsymbol{a} \mid \widetilde{A}}\right]=\left[\frac{1 \mid}{\ell \mid \widetilde{L}}\right]\left[\begin{array}{ll} \delta & \widetilde{D} \end{array}\right]\left[\frac{1 \mid \ell^T}{\mid \widetilde{L}^T}\right]=L D L^T .$$

$$\delta=\alpha, \delta \boldsymbol{\ell} \quad=\boldsymbol{a}, \quad \text { so } \boldsymbol{\ell}=\boldsymbol{a} / \delta=\boldsymbol{a} / \alpha \tilde{A}-\alpha \ell \ell^T=\widetilde{L} \widetilde{D} \widetilde{L}^T \quad\left(\text { recursive } L D L^T\right. \text { f }$$

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