# 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|DEFINITION AND EXAMPLES

Every algebraic structure has its corresponding functions (maps, morphisms). For vector spaces these functions are called linear transformations. The inputs are the vectors of one vector space and the outputs are the vectors from another (or perhaps the same) vector space. Just as subspaces are special subsets of a vector space, linear transformations are special functions from one vector space to another; they have the property that they respect (or preserve) the operations of the vector space. We give the formal definition.

Definition 4.1 Let $V$ and $W$ be two vector spaces. A function $T: V \longrightarrow W$ is a linear transformation if

1. For all $v_1, v_2 \in V$, we have $T\left(v_1+v_2\right)=T\left(v_1\right)+T\left(v_2\right)$.
2. For all scalars $a$ and $v \in V$, we have $T(a v)=a T(v)$.

We remark that the +’s in part a. refer to two different vector additions: $T\left(v_1+v_2\right)$ refers to addition in $V$ and $T\left(v_1\right)+T\left(v_2\right)$ refers to addition in $W$. There really is no reason for confusion since the context of the addition tells you which addition it must be. A similar remark can be made for part b. of the definition regarding scalar mulliplication.

We now list a few examples of linear transformations to make the reader more accustomed with the definition and also to illustrate the method by which one verifies that a function is indeed a linear transformation.

Example 4.1 Consider the function $T: \mathbb{R}^2 \longrightarrow P_1$ defined by $T([a, b])=b+a x$. For instance, according to its definition, $T([-1,3])=3-x$. In order to simplify our notation, we write just $T[a, b]$ instead of $T([a, b])$. We show that $T$ is a linear transformation by verifying parts $a$. and $b$. of the definition. To prove part a, notice that
$$\begin{gathered} T\left(\left[a_1, b_1\right]+\left[a_2, b_2\right]\right)=T\left[a_1+a_2, b_1+b_2\right]=\left(b_1+b_9\right)+\left(a_1+a_2\right) x \ =\left(b_1+a_1 x\right)+\left(b_2+a_2 x\right)=T\left[a_1, b_1\right]+T\left[a_2, b_2\right] . \end{gathered}$$
To prove part b, notice that
$$\begin{gathered} T\left(a\left[a_1, b_1\right]\right)=T\left[a a_1, a b_1\right]=\left(a b_1\right)+\left(a a_1\right) x \ =a\left(b_1+a_1 x\right)=a T\left[a_1, b_1\right] . \end{gathered}$$

## 数学代写|线性代数代写linear algebra代考|SIMILARITY OF MATRICES

In this section, we introduce the notion of similarity of square matrices. We show that the matrix representations of a linear operator have the special property of being similar to each other.

Definition 4.10 A matrix $A \in M_{n n}$ is similar (or conjugate) to a matrix $B \in$ $M_{n n}$ if there exists an invertible matrix $P \in M_{n n}$ such that $B=P^{-1} A P$.
Example $4.25$ Consider the following matrices:
$$A=\left[\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right] \text { und } B=\left[\begin{array}{rr} -1 & -1 \ 4 & 6 \end{array}\right]$$
Notice that $A$ is similar to $B$ since for invertible $P=\left[\begin{array}{ll}1 & 1 \ 1 & 2\end{array}\right]$ (invertible, since $|P|=1 \neq 0)$
$$P^{-1} A P=\left[\begin{array}{rr} 2 & -1 \ -1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \ 1 & 2 \end{array}\right]=\left[\begin{array}{rr} -1 & -1 \ 4 & 6 \end{array}\right]=B$$
Lemma 4.7 Let $A, B, C \in M_{n n}$. The following are true:
i. $A$ is similar to $A$.
ii. If $A$ is similar to $B$, then $B$ is similar to $A$.
iii. If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.
Proof 4.14 To prove $i$, take $P=I$. To prove ii, we are assuming that there is un inverlible $P$ such thal $P^{-1} A P=B$. Solving for $A$ we luve $A=P B P^{-1}=$ $\left(P^{-1}\right)^{-1} B P^{-1}$ and so the invertible matrix $P^{-1}$ makes $B$ similar to $A$. We leave the proof of iii as an exercise.

# 抽象代数代考

## 数学代写|线性代数代写线性代数代考|

For all $v_1, v_2 \in V$，我们有 $T\left(v_1+v_2\right)=T\left(v_1\right)+T\left(v_2\right)$.对于所有标量 $a$ 和 $v \in V$，我们有 $T(a v)=a T(v)$.

$$\begin{gathered} T\left(\left[a_1, b_1\right]+\left[a_2, b_2\right]\right)=T\left[a_1+a_2, b_1+b_2\right]=\left(b_1+b_9\right)+\left(a_1+a_2\right) x \ =\left(b_1+a_1 x\right)+\left(b_2+a_2 x\right)=T\left[a_1, b_1\right]+T\left[a_2, b_2\right] . \end{gathered}$$

$$\begin{gathered} T\left(a\left[a_1, b_1\right]\right)=T\left[a a_1, a b_1\right]=\left(a b_1\right)+\left(a a_1\right) x \ =a\left(b_1+a_1 x\right)=a T\left[a_1, b_1\right] . \end{gathered}$$

## 数学代写|线性代数代写线性代数代考|SIMILARITY OF MATRICES

4.10矩阵 $A \in M_{n n}$ 是否与矩阵相似(或共轭) $B \in$ $M_{n n}$ 如果存在一个可逆矩阵 $P \in M_{n n}$ 如此这般 $B=P^{-1} A P$.

$$A=\left[\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right] \text { und } B=\left[\begin{array}{rr} -1 & -1 \ 4 & 6 \end{array}\right]$$
$A$ 类似于 $B$ 对于可逆的 $P=\left[\begin{array}{ll}1 & 1 \ 1 & 2\end{array}\right]$ (可逆的，因为 $|P|=1 \neq 0)$
$$P^{-1} A P=\left[\begin{array}{rr} 2 & -1 \ -1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \ 1 & 2 \end{array}\right]=\left[\begin{array}{rr} -1 & -1 \ 4 & 6 \end{array}\right]=B$$

i。 $A$ 类似于 $A$.

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