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# 数学代写|线性代数代写Linear algebra代考|MS-E2121 Row and column vectors

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## 数学代写|线性代数代写Linear algebra代考|Row and column vectors

The elements of the vector space $\mathbb{K}^n$ are all the $n$-tuples of scalars from the field $\mathbb{K}$. There are two different ways that we can represent an $n$-tuple: as a row, or as a column. Thus, the vector with components 1,2 and $-3$ can be represented as a row vector
$$\left[\begin{array}{lll} 1 & 2 & -3 \end{array}\right]$$
or as a column vector
$$\left[\begin{array}{c} 1 \ 2 \ -3 \end{array}\right]$$
(Note that we use square brackets, rather than round brackets or parentheses. But you will see the notation $(1,2,-3)$ and the equivalent for columns in other books!) Both systems are in common use, and you should be familiar with both. The choice of row or column vectors makes some technical differences in the statements of the theorems, so care is needed.

There are arguments for and against both systems. Those who prefer row vectors would argue that we already use $(x, y)$ or $(x, y, z)$ for the coordinates of a point in 2- or 3-dimensional Euclidean space, so we should use the same for vectors. The most powerful argument will appear when we consider representing linear maps by matrices.

Those who prefer column vectors point to the convenience of representing, say, the linear equations
\begin{aligned} & 2 x+3 y=5 \ & 4 x+5 y=9 \end{aligned}
in matrix form
$$\left[\begin{array}{ll} 2 & 3 \ 4 & 5 \end{array}\right]\left[\begin{array}{l} x \ y \end{array}\right]=\left[\begin{array}{l} 5 \ 9 \end{array}\right]$$

## 数学代写|线性代数代写Linear algebra代考|Change of basis

The coordinate representation of a vector is always relative to a basis. We now have to look at how the representation changes when we use a different basis.
Definition 1.6 Let $B=\left(v_1, \ldots, v_n\right)$ and $B^{\prime}=\left(v_1^{\prime}, \ldots, v_n^{\prime}\right)$ be bases for the $n$-dimensional vector space $V$ over the field $\mathbb{K}$. The transitition matrix $P$ from $B$ to $B^{\prime}$ is the $n \times n$ matrix whose $j$ th column is the coordinate representation $\left[v_j^{\prime}\right]B$ of the $j$ th vector of $B^{\prime}$ relative to $B$. If we need to specify the bases, we write $P{B, B^{\prime}}$.

Proposition 1.6 Let $B$ and $B^{\prime}$ be bases for the n-dimensional vector space $V$ over the field $\mathbb{K}$. Then, for any vector $v \in V$, the coordinate representations of $v$ with respect to $B$ and $B^{\prime}$ are related by
$$[v]B=P[v]{B^{\prime}} .$$
Proof Let $p_{i j}$ be the $i, j$ entry of the matrix $P$. By definition, we have
$$v_j^{\prime}=\sum_{i=1}^n p_{i j} v_i$$
Take an arbitrary vector $v \in V$, and let
$$[v]B=\left[c_1, \ldots, c_n\right]^{\top}, \quad[v]{B^{\prime}}=\left[d_1, \ldots, d_n\right]^{\top} .$$
This means, by definition, that
$$v=\sum_{i=1}^n c_i v_i=\sum_{j=1}^n d_j v_j^{\prime}$$
Substituting the formula for $v_j^{\prime}$ into the second equation, we have
$$v=\sum_{j=1}^n d_j\left(\sum_{i=1}^n p_{i j} v_i\right)$$
Reversing the order of summation, we get
$$v=\sum_{i=1}^n\left(\sum_{j=1}^n p_{i j} d_j\right) v_i$$
Now we have two expressions for $v$ as a linear combination of the vectors $v_i$. By the uniqueness of the coordinate representation, they are the same: that is,
$$c_i=\sum_{j=1}^n p_{i j} d_j$$

## 数学代写|线性代数代写Linear algebra代考|Row and column vectors

$$\left[\begin{array}{lll} 1 & 2 & -3 \end{array}\right]$$

$$\left[\begin{array}{lll} 1 & 2 & -3 \end{array}\right]$$
（请注意，我们使用方括号，而不是圆括号或圆括号。但是您会看到符号 $(1,2,-3)$ 以及其他书中专栏的等效内容!）这两种系统 都很常用，您应该致悉它们。行或列向量的选择在定理的陈述中产生了一些技术上的差异，因此需要小心。

$$2 x+3 y=5 \quad 4 x+5 y=9$$

$$\left[\begin{array}{lll} 2 & 34 & 5 \end{array}\right]\left[\begin{array}{l} x y \end{array}\right]=\left[\begin{array}{l} 59 \end{array}\right]$$

## 数学代写|线性代数代写Linear algebra代考|Change of basis

$$[v] B=P[v] B^{\prime} .$$

$$v_j^{\prime}=\sum_{i=1}^n p_{i j} v_i$$

$$[v] B=\left[c_1, \ldots, c_n\right]^{\top}, \quad[v] B^{\prime}=\left[d_1, \ldots, d_n\right]^{\top} .$$

$$v=\sum_{i=1}^n c_i v_i=\sum_{j=1}^n d_j v_j^{\prime}$$

$$v=\sum_{j=1}^n d_j\left(\sum_{i=1}^n p_{i j} v_i\right)$$

$$v=\sum_{i=1}^n\left(\sum_{j=1}^n p_{i j} d_j\right) v_i$$

$$c_i=\sum_{j=1}^n p_{i j} d_j$$

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