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# 数学代写|数值分析代写Numerical analysis代考|MATH2722 Eigenvalues and Singular Values

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## 数学代写数值分析代写Numerical analysis代考|Eigenvalues and Eigenvectors

We inaugurate our discussion of eigenvalues and eigenvectors with the basic definition.
Definition 6.1. Let $A$ be an $n \times n$ matrix. A scalar $\lambda$ is called an eigenvalue of $A$ if there is a non-zero vector $\mathbf{v} \neq \mathbf{0}$, called an eigenvector, such that
$$A \mathbf{v}=\lambda \mathbf{v} \text {. }$$
In other words, the matrix $A$ stretches the eigenvector $\mathbf{v}$ by an amount specified by the eigenvalue $\lambda$.

Remark: The odd-looking terms “eigenvalue” and “eigenvector” are hybrid GermanEnglish words. In the original German, they are Eigenwert and Eigenvektor, which can be fully translated as “proper value” and “proper vector”. For some reason, the halftranslated terms have acquired a certain charm, and are now standard. The alternative English terms characteristic value and characteristic vector can be found in some (mostly older) texts. Oddly, the term characteristic equation, to be defined below, is still used.
The requirement that the eigenvector $\mathbf{v}$ be nonzero is important, since $\mathbf{v}=\mathbf{0}$ is a trivial solution to the eigenvalue equation (6.1) for any scalar $\lambda$. Moreover, as far as solving linear ordinary differential equations goes, the zero vector $\mathbf{v}=\mathbf{0}$ gives $\mathbf{u}(t) \equiv \mathbf{0}$, which is certainly a solution, but one that we already knew.

The eigenvalue equation (6.1) is a system of linear equations for the entries of the eigenvector $\mathbf{v}$ – provided that the eigenvalue $\lambda$ is specified in advance – but is “mildly” nonlinear as a combined system for $\lambda$ and $\mathbf{v}$. Gaussian Elimination per se will not solve the problem, and we are in need of a new idea. Let us begin by rewriting the equation in the form
$$(A-\lambda \mathrm{I}) \mathbf{v}=\mathbf{0},$$
where I is the identity matrix of the correct size ${ }^{\dagger}$. Now, for given $\lambda$, equation (6.2) is a homogeneous linear system for $\mathbf{v}$, and always has the trivial zero solution $\mathbf{v}=\mathbf{0}$. But we are specifically seeking a nonzero solution! A homogeneous linear system has a nonzero solution $\mathbf{v} \neq \mathbf{0}$ if and only if its coefficient matrix, which in this case is $A-\lambda \mathrm{I}$, is singular. This observation is the key to resolving the eigenvector equation.

## 数学代写|数值分析代写Numerical analysis代考|Completeness

Most of the vector space bases that play a distinguished role in applications are assembled from the eigenvectors of a particular matrix. In this section, we show that the eigenvectors of any “complete” matrix automatically form a basis for $\mathbb{R}^n$ or, in the complex case, $\mathbb{C}^n$. In the following subsection, we use the eigenvector basis to rewrite the linear transformation determined by the matrix in a simple diagonal form. The most important cases – symmetric and positive definite matrices – will be treated in the following section.
The first task is to show that eigenvectors corresponding to distinct eigenvalues are automatically linearly independent.

Lemma 6.13. If $\lambda_1, \ldots, \lambda_k$ are distinct eigenvalues of the same matrix $A$, then the corresponding eigenvectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ are linearly independent.

Proof: The result is proved by induction on the number of eigenvalues. The case $k=1$ is immediate since an eigenvector cannot be zero. Assume that we know the result is valid for $k-1$ eigenvalues. Suppose we have a vanishing linear combination:
$$c_1 \mathbf{v}1+\cdots+c{k-1} \mathbf{v}{k-1}+c_k \mathbf{v}_k=\mathbf{0} .$$ Let us multiply this equation by the matrix $A$ : \begin{aligned} A\left(c_1 \mathbf{v}_1+\cdots\right. & \left.+c{k-1} \mathbf{v}{k-1}+c_k \mathbf{v}_k\right)=c_1 A \mathbf{v}_1+\cdots+c{k-1} A \mathbf{v}{k-1}+c_k A \mathbf{v}_k \ & =c_1 \lambda_1 \mathbf{v}_1+\cdots+c{k-1} \lambda_{k-1} \mathbf{v}{k-1}+c_k \lambda_k \mathbf{v}_k=\mathbf{0} . \end{aligned} On the other hand, if we multiply the original equation (6.15) by $\lambda_k$, we also have $$c_1 \lambda_k \mathbf{v}_1+\cdots+c{k-1} \lambda_k \mathbf{v}{k-1}+c_k \lambda_k \mathbf{v}_k=\mathbf{0} .$$ Subtracting this from the previous equation, the final terms cancel and we are left with the equation $$c_1\left(\lambda_1-\lambda_k\right) \mathbf{v}_1+\cdots+c{k-1}\left(\lambda_{k-1}-\lambda_k\right) \mathbf{v}{k-1}=\mathbf{0} .$$ This is a vanishing linear combination of the first $k-1$ eigenvectors, and so, by our induction hypothesis, can only happen if all the coefficients are zero: $$c_1\left(\lambda_1-\lambda_k\right)=0, \quad \ldots \quad c{k-1}\left(\lambda_{k-1}-\lambda_k\right)=0 .$$
The eigenvalues were assumed to be distinct, so $\lambda_j \neq \lambda_k$ when $j \neq k$. Consequently, $c_1=\cdots=c_{k-1}=0$. Substituting these values back into (6.15), we find $c_k \mathbf{v}_k=\mathbf{0}$, and so $c_k=0$ also, since the eigenvector $\mathbf{v}_k \neq \mathbf{0}$. Thus we have proved that (6.15) holds if and only if $c_1=\cdots=c_k=0$, which implies the linear independence of the eigenvectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$. This completes the induction step.

## 数学代写数值分析代写Numerical analysis代考|Eigenvalues and Eigenvectors

$$A \mathbf{v}=\lambda \mathbf{v} .$$

$$(A-\lambda \mathrm{I}) \mathbf{v}=\mathbf{0},$$

## 数学代写|数值分析代写Numerical analysis代考|Completeness

$$c_1\left(\lambda_1-\lambda_k\right)=0, \quad \ldots \quad c k-1\left(\lambda_{k-1}-\lambda_k\right)=0 .$$

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