# 数学代写|代数学代写Algebra代考|MAT523

## 数学代写|代数学代写Algebra代考|A Characterization of Invertible Matrices

Since we now know how to find the inverse of a product of invertible matrices, it follows that if we can write a matrix as a product of elementary matrices, then we can find its inverse by inverting each of those elementary matrices and multiplying together those inverses in opposite order (just like we did in Example 2.2.6).

However, this observation is not yet useful, since we do not know how to break down a general invertible matrix into a product of elementary matrices, nor do we know whether or not this is always possible. The following theorem and its proof solve these problems, and also introduces an important connection between invertible matrices and systems of linear equations.
Suppose $A \in \mathcal{M}_n$. The following are equivalent:
a) $A$ is invertible.
b) The linear system $A \mathbf{x}=\mathbf{b}$ has a solution for all $\mathbf{b} \in \mathbb{R}^n$.
c) The linear system $A \mathbf{x}=\mathbf{b}$ has a unique solution for all $\mathbf{b} \in \mathbb{R}^n$.
d) The linear system $A \mathbf{x}=\mathbf{0}$ has a unique solution.
e The reduced row echelon form of $A$ is $I$ (the identity matrix).
f $A$ can be written as a product of elementary matrices.
Proof. To start, we show that (a) $\Longrightarrow$ (c) $\Longrightarrow$ (d) $\Longrightarrow$ (e) $\Longrightarrow$ (f) $\Longrightarrow$ (a), which means that any one of these five statements implies any of the other five (we will take care of condition (b) later).

To see that (a) $\Longrightarrow$ (c), we note that if $A$ is invertible then $\mathbf{x}=A^{-1} \mathbf{b}$ is a solution of the linear system $A \mathbf{x}=\mathbf{b}$, since
$$A\left(A^{-1} \mathbf{b}\right)=\left(A A^{-1}\right) \mathbf{b}=I \mathbf{b}=\mathbf{b} .$$
To see that this solution is unique, suppose that there were two solutions $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$. Then $A \mathbf{x}=\mathbf{b}$ and $A \mathbf{y}=\mathbf{b}$, so subtracting gives us $A(\mathbf{x}-\mathbf{y})=\mathbf{0}$. It then follows that
$$\mathbf{x}-\mathbf{y}=\left(A^{-1} A\right)(\mathbf{x}-\mathbf{y})=A^{-1}(A(\mathbf{x}-\mathbf{y}))=A^{-1} \mathbf{0}=\mathbf{0},$$
so $\mathbf{x}=\mathbf{y}$ (i.e., the solution is unique).

## 数学代写|代数学代写Algebra代考|Subspaces

As a starting point, recall that linear systems can be interpreted geometrically as asking for the point(s) of intersection of a collection of lines or planes (depending on the number of variables involved). The following definition introduces subspaces, which can be thought of as any-dimensional analogues of lines and planes.

A subspace of $\mathbb{R}^n$ is a non-empty set $\mathcal{S}$ of vectors in $\mathbb{R}^n$ with the properties that
a) if $\mathbf{v}, \mathbf{w} \in \mathcal{S}$ then $\mathbf{v}+\mathbf{w} \in \mathcal{S}$, and
b) if $\mathbf{v} \in \mathcal{S}$ and $c \in \mathbb{R}$ then $c \mathbf{v} \in \mathcal{S}$.
The idea behind this definition is that property (a) ensures that subspaces are “flat”, and property (b) makes it so that they are “infinitely long” (just like lines and planes). Subspaces do not have any holes or edges, and if they extend even a little bit in a given direction, then they must extend forever in that direction.

The defining properties of subspaces mimic the properties of lines and planes, but with one caveat-every subspace contains $\boldsymbol{0}$ (the zero vector). The reason for this is simply that if we choose $\mathbf{v} \in \mathcal{S}$ arbitrarily and let $c=0$ in property (b) of Definition $2.3 .1$ then
$$\mathbf{0}=0 \mathbf{v}=c \mathbf{v} \in \mathcal{S} .$$
This implies, for example, that a line through the origin is indeed a subspace, but a line in $\mathbb{R}^2$ with $y$-intercept equal to anything other than 0 is not a subspace (see Figure 2.13).

Before working with subspaces algebraically, we look at couple of quick examples geometrically to try to build some intuition for how they work.

# 代数学代考

## 数学代写|代数学代写Algebra代考|可逆矩阵的一个表征

a) $A$是可逆的
b)线性系统$A \mathbf{x}=\mathbf{b}$对所有的$\mathbf{b} \in \mathbb{R}^n$有一个解
c)线性系统$A \mathbf{x}=\mathbf{b}$对所有的$\mathbf{b} \in \mathbb{R}^n$有一个唯一解
d)线性系统$A \mathbf{x}=\mathbf{0}$有一个唯一解。
e $A$的行简化阶梯形为$I$(单位矩阵)。
f $A$可以写成初等矩阵的乘积。

$$A\left(A^{-1} \mathbf{b}\right)=\left(A A^{-1}\right) \mathbf{b}=I \mathbf{b}=\mathbf{b} .$$

$$\mathbf{x}-\mathbf{y}=\left(A^{-1} A\right)(\mathbf{x}-\mathbf{y})=A^{-1}(A(\mathbf{x}-\mathbf{y}))=A^{-1} \mathbf{0}=\mathbf{0},$$

## 数学代写|代数学代写Algebra代考|Subspaces

$\mathbb{R}^n$的子空间是一个非空集$\mathcal{S}$，由$\mathbb{R}^n$中的向量组成，其属性是
A)如果$\mathbf{v}, \mathbf{w} \in \mathcal{S}$则$\mathbf{v}+\mathbf{w} \in \mathcal{S}$，
b)如果$\mathbf{v} \in \mathcal{S}$和$c \in \mathbb{R}$则$c \mathbf{v} \in \mathcal{S}$。

$$\mathbf{0}=0 \mathbf{v}=c \mathbf{v} \in \mathcal{S} .$$

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