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## 数学代写|数值分析代写numerical analysis代考|The LU Decomposition with Pivoting

Gaussian elimination is referred to as a direct method for solving a linear system, meaning a method that terminates in finitely many iterations. These are very different from our methods of Ch. 1; in infinite precision arithmetic, Gaussian elimination would always give us the right answer in finitely many operations (like the quadratic formula does for quadratic equations), whereas Newton’s method would always give an approximation (even in infinite precision arithmetic). There are also iterative methods for linear systems that, like Newton’s method, do not reach their answer in finitely many steps, and that are useful for certain types of large systems (as we will discuss in Ch. 3). We’ll continue to focus on direct methods for now.

At this point our method for finding an LU decomposition works only when row-reducing the matrix does not require pivoting. There are types of matrices for which it can be shown that Gaussian elimination with partial pivoting will in fact never actually pivot (that is, interchange rows), but this is still a significant restriction. We need to remove it.

A matrix that does require pivoting need not have an LU decomposition at all. For example, consider the $2 \times 2$ matrix
$$\left[\begin{array}{ll} 0 & 1 \ 1 & 1 \end{array}\right] \text {. }$$
This is a square, nonsingular, well-behaved matrix. However, to have
$$\left[\begin{array}{ll} 0 & 1 \ 1 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \ m & 1 \end{array}\right]\left[\begin{array}{ll} a & b \ 0 & c \end{array}\right]$$
would require that (upon multiplying the matrices on the RHS) $a=0, b=1$, $a m=1$, and $m b+c=1$. But $a m=1$ is not possible, since $a=0$. Therefore this matrix does not admit an LU decomposition.

## 数学代写|数值分析代写numerical analysis代考|The Cholesky Decomposition

When we know something special about a matrix $A$ involved in a computational linear algebra problem we can often exploit it for additional efficiency. The two most commonly occurring cases are those in which we know that the matrix is sparse, that is, most of its entries are zeroes, and those in which we know that the matrix is symmetric, that is, that $A=A^T$. In this section we’ll focus on symmetric matrices that have an additional property: If $x$ is a nonzero vector then
$$x^T A x>0 .$$
A symmetric matrix with this property is said to be positive definite ${ }^8$. An equivalent characterization is that a matrix is positive definite if and only if it is symmetric and all of its eigenvalues are strictly positive. (Hence it will be nonsingular, as no eigenvalue is null.) A symmetric matrix with the property that $x^T A x \geq 0$ for all $x$ is said to be positive semi-definite and all its eigenvalues must be nonnegative. Similarly for negative definite $\left(x^T A x<0\right)$ and negative semi-definite $\left(x^T A x \leq 0\right)$ matrices.

Positive definite matrices appear more frequently than one might expect in practice. The quantity $x^T A x$ is called a quadratic form in $A$ and is the natural generalization of the scalar form $a x^2$; in this context it occurs in the generalization of Taylor series to functions of more than one variable where it is called the Hessian matrix. We’ll see in Ch. 7 that having the Hessian be positive definite is the natural analog of asking that $f^{\prime \prime}(x)$ be positive in single-variable calculus. This means that positive definite matrices appear often in multivariable minimization problems. Positive definite matrices also appear in least squares curve-fitting problems (see Sec. 2.7) and in the numerical solution of partial differential equations.

Since positive definite matrices are symmetric, they contain only about half as many distinct entries as a general matrix of the same order, and so we ought to be able to save about half the work and half the storage when dealing with them. However, an arbitrarily selected method will usually not take advantage of this structure. For example, the LU decomposition $A=L U$ of a symmetric matrix cannot have $L$ or $U$ symmetric in general ${ }^9$. We’ll focus on the more specific case of positive definite matrices. Here there is indeed a straight-forward procedure that produces an LU decomposition of the matrix in about half the time and using about half the storage, as we would expect.

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|The LU Decomposition with Pivoting

$$\left[\begin{array}{ll} 0 & 1 \ 1 & 1 \end{array}\right] \text {. }$$

$$\left[\begin{array}{ll} 0 & 1 \ 1 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \ m & 1 \end{array}\right]\left[\begin{array}{ll} a & b \ 0 & c \end{array}\right]$$

## 数学代写|数值分析代写数值分析代考| Cholesky分解

$$x^T A x>0 .$$

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## MATLAB代写

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