## 数学代写|线性代数代写linear algebra代考|Spectrum of a Matrix

Definition 33 Let $A$ be a square matrix in $\mathrm{M}_n(\mathbb{K})$. A non-zero vector $\boldsymbol{x} \in \mathbb{K}^n$ is said to be an eigenvector of $A$ if there exists $\lambda \in \mathbb{K}$ such that
$$A \mathrm{x}=\lambda \mathrm{x} .$$
Under these conditions, $\lambda$ is called an eigenvalue of $A$ associated with $\boldsymbol{x}$. The spectrum of $A$, denoted by $\sigma(A)$, is the set of eigenvalues of matrix $A$.

To find the spectrum of $A$, it is necessary to solve the equation (4.1) or, equivalently, to solve
$$(A-\lambda I) \mathbf{x}=\mathbf{0} .$$
Since we want to find $\lambda \in \mathbb{K}$ for which there exist non-zero vectors $\mathbf{x}$ satisfying (4.1), the homogeneous system (4.2) must have non-zero solutions. Hence, the coefficient matrix $A-\lambda I$ must be singular. By Theorem 2.1, this holds if and only if the determinant $\operatorname{det}(A-\lambda I)$ is equal to zero.

The polynomial $p(\lambda)=\operatorname{det}(A-\lambda I)$ has degree $n$ and is called the characteristic polinomyal of $A$. The equation
$$\operatorname{det}(A-\lambda I)=0$$
is the characteristic equation. The eigenvalues of matrix $A$ are, therefore, the roots of the characteristic polynomial of $A$.

Given an eigenvalue $\lambda$, the eigenspace $E(\lambda)$ associated with the eigenvalue $\lambda$ is the solution set of (4.2). In other words, $E(\lambda)$ is the null space of $A-\lambda I$, i.e.,
$$E(\lambda)=N(A-\lambda I)$$
Example 4.1 Find the spectrum and bases for the eigenspaces of
$$A=\left[\begin{array}{ccc} 0 & -1 & 0 \ -1 & 0 & 0 \ 0 & 0 & -1 \end{array}\right]$$

## 数学代写|线性代数代写linear algebra代考|Spectral Properties

Proposition 4.2 Let $A$ be a square matrix of order $n$ over $\mathbb{K}$, let
$$p(\lambda)=a_0+a_1 \lambda+\cdots+a_{n-1} \lambda^{n-1}+(-1)^n \lambda^n$$
be the characteristic polynomial of $A$, and let $\lambda_1, \ldots, \lambda_n$ be the eigenvalues of $A$ (possibly repeated). The following hold.
(i) $|A|=\lambda_1 \lambda_2 \cdots \lambda_n$.
(ii) $a_{n-1}=(-1)^{n-1} \operatorname{tr} A$.
(iii) $\operatorname{tr} A=\sum_{i=1}^n \lambda_i$.
Proof (i) Let
$$p(\lambda)=\left(\lambda_1-\lambda\right)\left(\lambda_2-\lambda\right) \cdots\left(\lambda_n-\lambda\right)$$
be the characteristic polynomial of $A$ (where the roots may not be all distinct).

Observing that
$$p(0)=|A-0 I|=|A|$$
and letting $\lambda=0$ in (4.4), we have
$$|A|=\lambda_1 \lambda_2 \cdots \lambda_n$$
(ii) This will be shown by induction. For $n=1$, the statement is clear. Suppose now that $n \geq 2$ and that the assertion holds for $n-1$. Let $A=\left[a_{i j}\right]$ be an $n \times n$ matrix and let $p(\lambda)=|A-\lambda I|$ be its characteristic polynomial. Hence, using the Laplace’s expansion along the first row, we have
$$p(\lambda)=\left(a_{11}-\lambda I\right) \operatorname{det}\left[(A-\lambda I){11}\right]+\sum{j=2}^n a_{1 j} C_{1 j} .$$

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Spectrum of a Matrix

$$A \mathrm{x}=\lambda \mathrm{x}$$

$$(A-\lambda I) \mathbf{x}=\mathbf{0} .$$

$$\operatorname{det}(A-\lambda I)=0$$

$$E(\lambda)=N(A-\lambda I)$$

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

## 数学代写|线性代数代写linear algebra代考|Spectral Properties

$$p(\lambda)=a_0+a_1 \lambda+\cdots+a_{n-1} \lambda^{n-1}+(-1)^n \lambda^n$$

\begin{aligned} & \text { (我) }|A|=\lambda_1 \lambda_2 \cdots \lambda_n . \ & \text { (二) } a_{n-1}=(-1)^{n-1} \operatorname{tr} A \text {. } \ & \text { (三) } \operatorname{tr} A=\sum_{i=1}^n \lambda_i . \end{aligned}

$$p(\lambda)=\left(\lambda_1-\lambda\right)\left(\lambda_2-\lambda\right) \cdots\left(\lambda_n-\lambda\right)$$

$$p(0)=|A-0 I|=|A|$$

$$|A|=\lambda_1 \lambda_2 \cdots \lambda_n$$
(ii) 这将通过归纳来证明。为了 $n=1$ ，说法很明确。现在假设 $n \geq 2$ 并且该断言适用于 $n-1$. 让 $A=\left[a_{i j}\right]$ 豆 $n \times n$ 矩阵并让 $p(\lambda)=|A-\lambda I|$ 是它的特征多项式。因此，沿 第一行使用拉普拉斯展开，我们有
$$p(\lambda)=\left(a_{11}-\lambda I\right) \operatorname{det}[(A-\lambda I) 11]+\sum j=2^n a_{1 j} C_{1 j}$$

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