# 数学代写|线性代数代写linear algebra代考|MTH2106

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

Suppose that $B$ is an $n \times n$ matrix. Did you notice that Proposition $4.9$ forces the existence of a subspace of vectors $\mathrm{x} \in \mathbb{K}^n$ to satisfy $B^r \mathbf{x}=\mathbf{0}$, for some non-negative integer $r$ which depends on $\mathbf{x}$ ? (We are referring to the subspace $N\left(B^n\right)$.) Surely it suffices to impose $r=n$. But we may find that this holds for $r<n$, depending on the vectors $\mathbf{x}$ in $N\left(B^n\right)$. Hence, given a non-zero $\mathrm{x} \in N\left(B^n\right)$, should it exist, there is a smallest $r$ such that
$$B^r \mathbf{x}=\mathbf{0} \text { and } B^{r-1} \mathbf{x} \neq \mathbf{0} .$$
Hold this thought when reading the next definition.
Definition 42 Let $A$ be an $n \times n$ matrix over $\mathbb{K}$ and let $\lambda \in \mathbb{K}$ be an eigenvalue of $A$. A non-zero vector $\mathrm{x} \in \mathbb{K}^n$ is said to be a generalised eigenvector of A associated with $\lambda$ if there exists $k \in \mathbb{N}$ such that
$$(A-\lambda I)^k \mathbf{x}=\mathbf{0} .$$
The order of the generalised eigenvector $\mathbf{x}$ is the smallest $k \in \mathbb{N}$ such that $\mathrm{x} \in N(A-\lambda I)^k$.

Notice that (4.8) is the same as requiring that $\mathbf{x}$ lie in the null space $N(A-\lambda I)^k$ of the matrix $(A-\lambda I)^k$ and, consequently, in all the null spaces of higher powers of $A-\lambda I$. Hence the definition of the order of the generalised eigenvector as the smallest $k$ for which (4.8) holds. In other words, $\mathbf{x}$ is a generalised eigenvector of order $k$ of $A$, associated with $\lambda$, if
$$(A-\lambda I)^k \mathbf{x}=\mathbf{0} \quad \text { and } \quad(A-\lambda I)^{k-1} \mathbf{x} \neq \mathbf{0} .$$

## 数学代写|线性代数代写linear algebra代考|Jordan canonical form

As seen in Example 4.6, we found a basis of $\mathbb{R}^3$ consisting only of generalised eigenvectors. In particular, the vectors in each of the Jordan chains in the example were linearly independent. In turns out that this is a general result.

Proposition $4.15$ Let $A$ bc a matrix in $\mathrm{M}n(\mathbb{K})$, lct $\lambda \in \mathbb{K} b c$ an cigcnvaluc of $A$ and let $\mathrm{x}$ be a generalised eigenvector of order $k$ of $A$ associated with the eigenvalue $\lambda$. Then, the vectors in the Jordan chain $$(A-\lambda I)^{k-1} \mathbf{x}, \quad(A-\lambda I)^{k-2} \mathbf{x}, \quad \ldots, \quad(A-\lambda I) \mathbf{x}, \quad \mathbf{x}$$ are linearly independent. Proof Let $k$ and $\mathbf{x}$ be as above and let $$\underbrace{(A-\lambda I)^{k-1} \mathbf{x}}{\mathbf{u}1}, \underbrace{(A-\lambda I)^{k-2} \mathbf{x}}{\mathbf{u}2}, \cdots \underbrace{(A-\lambda I) \mathbf{x}}{\mathbf{u}{k-1}}, \underbrace{\mathbf{x}}{\mathbf{u}_k} .$$
If $k=1$, then the set $\left{\mathbf{u}_1\right}$ is linearly independent, since $\mathbf{u}_1 \neq 0$. Hence, the proposition is proved for $k=1$.

Let now $k>1$, let $p$ be an integer such that $1 \leq p<k$ and consider the set $S_p=\left{\mathbf{u}1, \ldots, \mathbf{u}_p\right}$. Observe that, if $p=1$, it can be shown similarly to the above paragraph that $S_1$ is linearly independent. We wish to show that, fixing $p$ and assuming that $S_p$ is linearly independent, then $$S{p+1}=\left{\mathbf{u}1, \ldots, \mathbf{u}_p, \mathbf{u}{p+1}\right}$$
is also linearly independent.

# 线性代数代考

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

$$B^r \mathbf{x}=\mathbf{0} \text { and } B^{r-1} \mathbf{x} \neq \mathbf{0} .$$

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

$$(A-\lambda I)^k \mathbf{x}=\mathbf{0} \quad \text { and } \quad(A-\lambda I)^{k-1} \mathbf{x} \neq \mathbf{0}$$

## 数学代写|线性代数代写linear algebra代考|Jordan canonical form

$$(A-\lambda I)^{k-1} \mathbf{x}, \quad(A-\lambda I)^{k-2} \mathbf{x}, \quad \cdots, \quad(A-\lambda I) \mathbf{x}, \quad \mathbf{x}$$

$$\underbrace{(A-\lambda I)^{k-1} \mathbf{x}} \mathbf{u} 1, \underbrace{(A-\lambda I)^{k-2} \mathbf{x}} \mathbf{u} 2, \cdots \underbrace{(A-\lambda I) \mathbf{x}} \mathbf{u} k-1, \underbrace{\mathbf{x}} \mathbf{u}k .$$ 如果 $k=1$ ，那么集合 \eft{mathbf{u}_1\right } } \text { 是线性独立的，因为 } \mathbf { u } { 1 } \neq 0 \text { . 因此，命题被证明为 } k = 1 \text { . }

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