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

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 $A$ be a matrix in $\mathrm{M}n(\mathbb{K})$, let $\lambda \in \mathbb{K}$ be an eigenvalue 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 $$(A-\lambda I)^{k-1} \mathbf{x}, \quad(A-\lambda I)^{k-2} \mathbf{x}, \quad \cdots, \quad(A-\lambda I) \mathbf{x}, \quad \mathbf{x}$$ 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.

数学代写|线性代数代写linear algebra代考|Linear Transformations

Definition 43 Let $U$ and $V$ be vector spaces over $\mathbb{K}$. A function $T: U \rightarrow V$ is called a linear transformation if, for all $\boldsymbol{x}, \boldsymbol{y} \in U$ and $\alpha \in \mathbb{K}$,
\begin{aligned} T(\boldsymbol{x}+\boldsymbol{y}) & =T(\boldsymbol{x})+T(\boldsymbol{y}), \ T(\alpha \boldsymbol{x}) & =\alpha T(\boldsymbol{x}) \end{aligned}
In other words, a function $T: U \rightarrow V$ is a linear transformation if it is additive (5.1) and homogeneous (5.2). As usual, $U$ is the domain of $T$ and $V$ is its codomain.

Proposition 5.1 Let $U$ and $V$ be vector spaces over $\mathbb{K}$ and let $T: U \rightarrow V$ be a linear transformation. Then
$$T\left(\mathbf{0}_U\right)=\mathbf{0}_V$$
where $\mathbf{0}_U$ is the zero vector in $U$ and $\mathbf{0}_V$ is the zero vector in $V$.
Proof Let $\boldsymbol{x}$ be a vector in $U$. Hence, by (5.1),
$$T(\boldsymbol{x})=T\left(\boldsymbol{x}+\mathbf{0}_U\right)=T(\boldsymbol{x})+T\left(\mathbf{0}_U\right)$$
from which follows that $T\left(\mathbf{0}_U\right)=\mathbf{0}_V$.
Example 5.1 Find which of the following functions are linear transformations.
a) $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a reflection relative to the $x$-axis.
b) $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ is an orthogonal projection on the xy-plane.
c) $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a translation by the vector $\boldsymbol{u}=(1,0)$.
The function $T$ in a) is defined, for all $\boldsymbol{x}=\left(x_1, x_2\right)$ in $\mathbb{R}^2$, by
$$T\left(x_1, x_2\right)=\left(x_1,-x_2\right)$$

线性代数代考

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

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

$$\underbrace{(A-\lambda I)^{k-1} \mathbf{x}} \mathbf{u} 1,(\underbrace{(A-\lambda I)^{k-2} \mathbf{x}} \mathbf{u} 2, \cdots \underbrace{(A-\lambda I) \mathbf{x}} \mathbf{u} k-1, \underbrace{\mathbf{x}} \mathbf{u}_k .$$

$S{p+1}=\backslash$ left $\backslash$ mathbf{u}1, $\backslash$ Idots, $\left.\backslash m a t h b f{u} _p, \backslash m a t h b f{u}{p+1} \backslash \mid i g h t\right}$

数学代写|线性代数代写linear algebra代考|Linear Transformations

$$T(\boldsymbol{x}+\boldsymbol{y})=T(\boldsymbol{x})+T(\boldsymbol{y}), T(\alpha \boldsymbol{x}) \quad=\alpha T(\boldsymbol{x})$$

$$T\left(\mathbf{0}_U\right)=\mathbf{0}_V$$

$$T(\boldsymbol{x})=T\left(\boldsymbol{x}+\mathbf{0}_U\right)=T(\boldsymbol{x})+T\left(\mathbf{0}_U\right)$$

A) $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ 是相对于 $x$-轴。

C) $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ 是向量的翻译 $\boldsymbol{u}=(1,0)$.

$$T\left(x_1, x_2\right)=\left(x_1,-x_2\right)$$

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