##### 以下是MATH 4260 – Numerical Analysis: Linear and Nonlinear Problems的题目的简单解析
Problem 1.

Let
$$A=\left[\begin{array}{rrr} 0 & -2 & 1 \\ 1 & -1 & 3 \\ -2 & 1 & -4 \end{array}\right] \quad \text { and } \quad b=\left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right]$$
(a) [5 Points] Apply LU factorization with partial pivoting to $A$.

(b) [4 Points] Use your results obtained in (a) to solve the linear system $A x=b$.

Proof .

Solution: In the first step, we need to interchange rows 1 and 3:
$$P=\left[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right], \quad\left[\begin{array}{rrr} -2 & 1 & -4 \\ 1 & -1 & 3 \\ 0 & -2 & 1 \end{array}\right] \quad \longrightarrow \quad\left[\begin{array}{rrr} -2 & 1 & -4 \\ -\frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & -2 & 1 \end{array}\right]$$
In the second step, we need to interchange rows 2 and 3 :
$$\left.P=\left[\begin{array}{lll} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right], \quad\left[\begin{array}{rrr} -2 & 1 & -4 \\ 0 & -2 & 1 \\ -\frac{1}{2} & -\frac{1}{2} & 1 \end{array}\right] \stackrel{0}{2}\right] \stackrel{0}{\longrightarrow}\left[\begin{array}{rrr} -2 & 1 & -4 \\ 0 & -2 & 1 \\ -\frac{1}{2} & \frac{1}{4} & \frac{3}{4} \end{array}\right]$$
Thus
$$P=\left[\begin{array}{lll} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right], \quad L=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{1}{2} & \frac{1}{4} & 1 \end{array}\right], \quad \text { and } \quad U=\left[\begin{array}{rrr} -2 & 1 & -4 \\ 0 & -2 & 1 \\ 0 & 0 & \frac{3}{4} \end{array}\right]$$

Since $P A=L U$, solving $A x=b$ is equivalent to solving the lower-triangular linear system $L y=P b$ for $y$ and the upper-triangular linear system $U x=y$ for $x$.
Solution of $L y=P b$ :
$$\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{1}{2} & \frac{1}{4} & 1 \end{array}\right] y=P b=\left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right] \quad \Longrightarrow \quad y=\left[\begin{array}{r} -1 \\ 1 \\ -\frac{3}{4} \end{array}\right]$$
Solution of $U x=y$ :
$$\left[\begin{array}{rrr} -2 & 1 & -4 \\ 0 & -2 & 1 \\ 0 & 0 & \frac{3}{4} \end{array}\right] x=\left[\begin{array}{r} -1 \\ 1 \\ -\frac{3}{4} \end{array}\right] \quad \Longrightarrow \quad x=\left[\begin{array}{r} 2 \\ -1 \\ -1 \end{array}\right]$$

Problem 2.

Let $A \in \mathbb{R}^{6 \times 6}$ be a matrix with characteristic polynomial
$$p_{A}(\lambda):=\operatorname{det}(A-\lambda I)=(\lambda+4)(\lambda-1)(\lambda-2)(\lambda-3)\left(\lambda^{2}+1\right)$$
(a) [2 Points] Determine all eigenvalues of $A$.

(b) [3 Points] We apply the power method to $A$. Which eigenvalue does the method converge to and what is the the ratio of the convergence rate?

(c) [4 Points] We apply inverse iteration with shift $s=\frac{7}{4}$ to $A$. Which eigenvalue does the method converge to and what is the ratio of the convergence rate?
Solution: The eigenvalues of the shifted and inverted matrix

Proof .

Solution: The eigenvalues of $A$ are the zeros of $p_{A}(\lambda):$
$$-4,1,2,3, i,-i$$
where $i=\sqrt{-1}$

The dominant eigenvalue of $A$ is $-4,$ and the second largest in absolute value is $3 .$ Thus the power method converges to the eigenvalue -4 and
$$\text { ratio }=\frac{3}{4}$$

The eigenvalues of the shifted and inverted matrix
$$(A-s I)^{-1}=\left(A-\frac{7}{4} I\right)^{-1}$$
are
$$\frac{1}{-4-\frac{7}{4}}, \quad \frac{1}{1-\frac{7}{4}}, \quad \frac{1}{2-\frac{7}{4}}, \quad \frac{1}{3-\frac{7}{4}}, \quad \frac{1}{1-\frac{7}{4}}, \quad \frac{1}{-\mathrm{i}-\frac{7}{4}}$$
The dominant eigenvalue of $(A-s I)^{-1}$ is
$$\frac{1}{2-\frac{7}{4}}=4$$
and the second largest in absolute value is
$$\frac{1}{1-\frac{7}{4}}=-\frac{4}{3}$$
Thus inverse iteration converges to the eigenvalue 2 of $A$ and
ratio $=\frac{1}{3}$.

Problem 3.

Let
$$A=\left[\begin{array}{rc} 1 & -1+\delta \\ -1 & 1 \end{array}\right]$$
where $\delta \in \mathbb{R}$ is a parameter.
(a) [2 Points] Show that $A$ is nonsingular if, and only if, $\delta \neq 0$. Determine $A^{-1}$ for $\delta \neq 0$.

(b) [4 Points] For $\delta \neq 0$, determine the condition number $\kappa_{1}(A)$ of $A$ with respect to the matrix norm $\|\cdot\|_{1}$

(c) [3 Points] Discuss the accuracy of the computed solution that you can expect when you solve linear systems $A x=b$ numerically.

Proof .

Solution: Since
$$\operatorname{det} A=1 \cdot 1-(-1+\delta) \cdot(-1)=\delta$$
the matrix $A$ is nonsingular if, and only if, $\delta \neq 0$.
For $\delta \neq 0$
$$A^{-1}=\frac{1}{\operatorname{det} A} \operatorname{Adj}(A)=\frac{1}{\delta}\left[\begin{array}{cc} 1 & 1-\delta \\ 1 & 1 \end{array}\right]$$

Recall that $\|\cdot\|_{1}$ is the maximumabsolute column sum of the matrix. Thus, we have
$$\|A\|_{1}=\max \{2,|-1+\delta|+1\}=\left\{\begin{array}{ll} 2 & \text { if } 0<\delta \leq 2 \\ 1+|\delta-1| & \text { otherwise } \end{array}\right.$$
and
$$\left\|A^{-1}\right\|_{1}=\frac{1}{|\delta|} \max \{2,|1-\delta|+1\}=\left\{\begin{array}{ll} \frac{2}{|\delta|} & \text { if } 0<\delta \leq 2, \\ \frac{1+|\delta-1|}{|\delta|} & \text { otherwise. } \end{array}\right.$$
It follows that
$$\kappa_{1}(A)=\|A\|_{1} \cdot\left\|A^{-1}\right\|_{1}=\left\{\begin{array}{ll} \frac{4}{|\delta|} & \text { if } 0<\delta \leq 2, \\ \frac{(1+|\delta-1|)^{2}}{|\delta|} & \text { otherwise } \end{array}\right.$$

If $\kappa_{1}(A) \gg 1,$ the problem of solving linear systems $A x=b$ is ill-conditioned. In this case, we cannot expect to solve $A x=b$ numerically with any reasonable accuracy, no matter what algorithm we employ. The expression for $\kappa_{1}(A)$ obtained in (b) shows that $A x=b$ is ill-conditioned for $\delta \approx 0$ and for $|\delta| \gg 1$.

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