# 数学代写|线性代数代写linear algebra代考|МАTH1014

## 数学代写|线性代数代写linear algebra代考|SUBSPACES ASSOCIATED WITH A MATRIX

In this section we address some results that were earlier discussed but never formally proved. This section also gives a justification for the process of finding a basis for a span of a set of vectors. We now remind the reader of some definitions as well as introduce some new notation and terminology.

Definition $3.18$ For $A \in M_{m n}$ with rows $r_1, \ldots, r_m \in \mathbb{R}^n$ and columns $c_1, \ldots, c_n \in$ $\mathbb{R}^m$

1. the row space of $A$, which we will denote by
$$\operatorname{rowsp}(A)=\operatorname{span}\left(r_1, \ldots, r_m\right) \text { a subspace of } \mathbb{R}^n .$$
2. the column space of $A$, which we will denote by
$$\operatorname{colsp}(\mathrm{A})=\operatorname{span}\left(c_1, \ldots, c_n\right) \text { a subspace of } \mathbb{R}^m .$$
3. the null space of $A$, which we will denote by
$$\text { nullsp(A) }=\left{u \in \mathbb{R}^n \mid A u=0\right} \text { a subspace of } \mathbb{R}^n \text {. }$$
4. the dimension of $\operatorname{rowsp}(A)$ will be called the rank of $A$ (for good reason) and will be denoted by $r(A)$.
5. the dimension of nullsp $(A)$ will be called the nullity of $A$ and will be denoted by $n(A)$.
We leave the proof of the following lemma as an exercise:
Lemma 3.8 If $A \in M_{m n}$ and $B \in \mathbb{R}^m$, then
i. $\operatorname{rowsp}\left(A^T\right)=\operatorname{colsp}(A)$ and $\operatorname{colsp}\left(A^T\right)=\operatorname{rowsp}(A)$.
ii. $\operatorname{colsp}(A)=\left{A v \mid v \in \mathbb{R}^n\right}$.
iii. The linear system $A X=B$ is consistent iff $B \in \operatorname{colsp}(A)$.

## 数学代写|线性代数代写linear algebra代考|APPLICATION: DIMENSION THEOREMS

In this section, we illustrate a technique called a counting argument. These counting arguments will be a consequence of the two dimension theorems proved in this section. The first dimension theorem is a result we would expect to be true if dimension is indeed measuring size in some sense. First, we need a lemma.
Lemma $3.12$ For $v_1, \ldots, v_n \in V$ a vector space, the following are equivalent:
i. The vectors $v_1, \ldots, v_n$ represent a maximal number of linearly independent vectors in $V$.
ii. $v_1, \ldots, v_n$ form a basis for $V$.
Proof $3.21$ To show that $i$. implies ii, We need to show that $v_1, \ldots, v_n$ span $V$. Suppose lo lhe conlrury llual lhey don’l spun $V$. This means lhere exisls a $v \in V$ mol in the span of $v_1, \ldots, v_n$. By Lemma $3.7, v_1, \ldots, v_n, v$ are linearly independent. But this contradicts the fact that $v_1, \ldots, v_n$ is a largest number of linearly independent vectors in $V$. Hence, it must be that case that $v_1, \ldots, v_n$ form a basis for $V$. The reverse implication is left as an exercise.

Theorem $3.13$ (Subspace Dimension Theorem) Let $V$ be a vector space with finite dimension and $U$ be a subspace of $V$. Then
i. $\operatorname{dim} U \leq \operatorname{dim} V$.
ii. $\operatorname{dim} U=\operatorname{dim} V$ iff $U=V$

Proof $3.22$ Let $v_1, \ldots, v_n$ be a basis for $V($ so $\operatorname{dim} V=n)$. To prove $i$, if $U={0}$ then certainly $\operatorname{dim} U=0 \leq \operatorname{dim} V$, so we can assume that $U \neq{0}$. Consider all collections $u_1, \ldots, u_k \in U$ which are linearly independent. Such collections exist since, for instance, a single $0 \neq u \in U$ is linearly independent. By Lemma 3.6, for every such collection we have $k \leq n$ (i.e. there is a bound on how large $k$ can be). Hence, there is a collection $u_1, \ldots, u_k$ linearly independent with $k$ largest. By Lemma 3.12, this largest collection forms a basis for $U$. Thus, $\operatorname{dim} U=k \leq n=\operatorname{dim} V$.

To prove ii, one direction is trivial: If $U=V$ then certainly $\operatorname{dim} U=\operatorname{dim} V$. Now lets assume that $\operatorname{dim} U=\operatorname{dim} V$. Choose $u_1, \ldots, u_n$ a basis for $U$. Since $u_1, \ldots, u_n$ are linearly independent in $V$, by Lemma 3.10, $u_1, \ldots, u_n$ is a basis for $V$, and so they span V. Thus,
$$V=\operatorname{span}\left(u_1, \ldots, u_n\right)=U$$

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## 数学代写|线性代数代写线性代数代考|SUBSPACES ASSOCIATED WITH A MATRIX

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