# 数学代写|线性代数代写linear algebra代考|MATH1014

## 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
$$p(\lambda)=\left(a_{11}-\lambda I\right) \operatorname{det}\left[(A-\lambda I){11}\right]+\sum{j=2}^n a_{1 j} C_{1 j} .$$
Recall that $\left[(A-\lambda I){11}\right]$ is the $n-1 \times n-1$ matrix obtained by deleting the first row and the first column of $A-\lambda I$. It then follows from the induction hypothesis that the coefficient corresponding to $\lambda^{n-1}$ is $$a{11}(-1)^{n-1}-(-1)^{n-2} \operatorname{tr}\left[A_{11}\right]=(-1)^{n-1} \operatorname{tr} A .$$
Assertion (iii) follows immediately from (4.4) and assertion (ii).
A useful fact to keep in mind is that, for an $n \times n$ matrix $A$ over $\mathbb{K}$ with characteristic polynomial
$$p(\lambda)=\left(\lambda_1-\lambda\right)^{r_1}\left(\lambda_2-\lambda\right)^{r_2} \cdots\left(\lambda_k-\lambda\right)^{r_k},$$
where $\lambda_1, \lambda_2, \ldots, \lambda_k$ are all distinct, we have
$$r_1+r_2+\cdots+r_k=n .$$
Before the next proposition, we need to make a definition.

## 数学代写|线性代数代写linear algebra代考|Similarity and Diagonalisation

We begin here the discussion of similarity mentioned at the beginnig of this chapter.

Definition 37 Let $A$ and $B$ be real (resp., complex) $n \times n$ matrices. $B$ is said to be similar to $A$ if there exists an invertible matrix $S$ such that
$$B=S^{-1} A S$$
or, equivalently, if
$$S B=A S .$$
It is easy to see that $B$ is similar to $A$ if and only if $A$ is similar to $B$. Consequently, to simplify one says simply that $A$ and $B$ are similar matrices.
Theorem 4.2 Let $A$ and $B$ be $n \times n$ similar matrices.
(i) $|A|=|B|$.
(ii) $A$ is invertible if and only if $B$ is invertible.
(iii) The characteristic polynomial $p_A(\lambda)$ of $A$ coincides with the characteristic polynomial $p_B(\lambda)$ of $B$.
(iv) $\operatorname{tr} A=\operatorname{tr} B$
(v) $\sigma(A)=\sigma(B)$ and the corresponding algebraic multiplicities (respectively, geometric multiplicities) of each eigenvalue coincide.
(vi) $\operatorname{dim} N(A)=\operatorname{dim} N(B)$.
(vii) $\operatorname{rank} A=\operatorname{rank} B$.
Proof (i) Since $A$ and $B$ are similar, we have $|B|=\left|S^{-1} A S\right|$. Hence, by Propositon 2.3 and Corollary 2.1,
$$|B|=\left|S^{-1}\right||A||S|=|A|\left|S^{-1}\right||S|=|A| .$$
(ii) By (i), we know that $|A| \neq 0$ if and only if $|B| \neq 0$. Hence, $A$ is invertible if and only if $B$ is invertible.
(iii) We have
$$p_B(\lambda)=|B-\lambda I|=\left|S^{-1} A S-\lambda I\right|=\left|S^{-1} A S-\lambda S^{-1} I S\right| .$$
Hence
$$p_B(\lambda)=\left|S^{-1}(A-\lambda I) S\right|=|A-\lambda I|=p_A(\lambda) .$$
(iv) This is a direct consequence of (iii) and Proposition $4.2$ (ii).

# 线性代数代考

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

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

$$\text { (二) } a_{n-1}=(-1)^{n-1} \operatorname{tr} A \text {. }$$
(三) $\operatorname{tr} A=\sum_{i=1}^n \lambda_i$

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

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

$$|A|=\lambda_1 \lambda_2 \cdots \lambda_n .$$
(ii) 这将通过归纳来证明。为了 $n=1$ ，陈述很清楚。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，使用沿第一行的拉普拉斯展开 式，我们有
$$p(\lambda)=\left(a_{11}-\lambda I\right) \operatorname{det}[(A-\lambda I) 11]+\sum j=2^n a_{1 j} C_{1 j} .$$

$$a 11(-1)^{n-1}-(-1)^{n-2} \operatorname{tr}\left[A_{11}\right]=(-1)^{n-1} \operatorname{tr} A .$$

$$p(\lambda)=\left(\lambda_1-\lambda\right)^{r_1}\left(\lambda_2-\lambda\right)^{r_2} \cdots\left(\lambda_k-\lambda\right)^{r_k},$$

$$r_1+r_2+\cdots+r_k=n .$$

## 数学代写|线性代数代写linear algebra代考|Similarity and Diagonalisation

$$B=S^{-1} A S$$

$$S B=A S .$$

(二) $A$ 当且仅当是可逆的 $B$ 是可逆的。
(iii) 特征多项式 $p_A(\lambda)$ 的 $A$ 符合特征多项式 $p_B(\lambda)$ 的 $B$.
(四) $\operatorname{tr} A=\operatorname{tr} B$
(在) $\sigma(A)=\sigma(B)$ 并且每个特征值对应的代数重数 (分别为几何重数) 重合。
(六) $\operatorname{dim} N(A)=\operatorname{dim} N(B)$.
(七) $\operatorname{rank} A=\operatorname{rank} B$.

$$|B|=\left|S^{-1}\right||A||S|=|A|\left|S^{-1}\right||S|=|A| .$$
(ii) 通过 (i)，我们知道 $|A| \neq 0$ 当且仅当 $|B| \neq 0$. 因此， $A$ 当且仅当是可逆的 $B$ 是可逆的。
(iii) 我们有
$$p_B(\lambda)=|B-\lambda I|=\left|S^{-1} A S-\lambda I\right|=\left|S^{-1} A S-\lambda S^{-1} I S\right| .$$

$$p_B(\lambda)=\left|S^{-1}(A-\lambda I) S\right|=|A-\lambda I|=p_A(\lambda) .$$

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