## 数学代写|线性代数代写linear algebra代考|Matrix spaces and spaces of polynomials

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## 数学代写|线性代数代写linear algebra代考|Matrix spaces and spaces of polynomials

Up to this point and although our definitons are general, in practical terms we have been focusing on $\mathbb{K}^n$. It is now the time to see what we get when addressing these concepts in more general vector spaces.

The ordered standard basis of the space $\mathrm{M}{k, n}(\mathbb{K})$ of the $k \times n$ real matrices is the ordered set consisting of the real $k \times n$ matrices having all entries but one equal to zero which takes value 1 ; the ordering is such that the non-zero entry in the first matrix is entry-11 and it ‘circulates’ along the lines from left to right. For example, in the case of $\mathrm{M}_2(\mathbb{K})$, the standard basis is $$\mathcal{B}_c=\left(\left[\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right]\right)$$ Given a matrix $A=\left[a{i j}\right]$, we have
$$A=\left[\begin{array}{ll} a_{11} & a_{12} \ a_{21} & a_{22} \end{array}\right]=a_{11}\left[\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right]+a_{12}\left[\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right]+a_{21}\left[\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right]+a_{22}\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right] .$$
Hence it is clear that $A$ is a linear combination of the vectors, i.e., of the matrices, in $\mathcal{B}c$. It is also easy to see that $\mathcal{B}_c$ is a linearly independent set. In fact $$\alpha_1\left[\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right]+\alpha_2\left[\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right]+\alpha_3\left[\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right]+\alpha_4\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \ 0 & 0 \end{array}\right],$$ yields $\alpha_1=\alpha_2=\alpha_3=\alpha_4=0$, which shows that $\mathcal{B}_c$ is linearly independent. We see that $\mathcal{B}_c$ is a basis of $\mathrm{M}_2(\mathbb{K})$ and that, for a matrix $A$ as above, the coordinate vector $A{\mathcal{B}c}$ of $A$ relative to the basis $\mathcal{B}_c$ is $\left(a{11}, a_{12}, a_{21}, a_{22}\right)$ which lies in $\mathbb{K}^4$. We have then that any matrix
$$A=\left[\begin{array}{ll} a_{11} & a_{12} \ a_{21} & a_{22} \end{array}\right]$$
has an image in $\mathbb{K}^4$ according to
\begin{aligned} T: \mathrm{M}2(\mathbb{K}) & \rightarrow \mathbb{K}^4 \ A & \mapsto A{\mathcal{B}}=\left(a_{11}, a_{12}, a_{21}, a_{22}\right) . \end{aligned}
Observe that Proposition 3.5 guarantees that $T$ is bijective.

## 数学代写|线性代数代写linear algebra代考|Existence and construction of bases

Lately we have seen how important bases are: they act as a ‘system of coordinates’ with respect to which the space is described. This even allows for treating any space having a basis with $n$ vectors like $\mathbb{K}^n$ (cf. §3.3.1).

One might ask however whether this is always possible. Given a space, does it always have a basis? And if it has two bases, say, is there a relation between their cardinality?

The next two theorems answer these questions for spaces having a spanning set. But before going into that, it should be pointed out that not all spaces have a spanning set, that is, a finite set whose span coincides with the space. For example, if one considers the set $\mathbb{P}$ of real polynomials, it is impossible to find such a set for $\mathbb{P}$. (Why?)
Theorem 3.3 Every vector space over $\mathbb{K}$ with a spanning set has a basis.
Here we adopt the convention that the empty set $\emptyset$ is a basis of $V={\mathbf{0}}$.
Proof The case $V={0}$ holds trivially. Let $V \neq{0}$ and let $X$ be a spanning set of $V$. We show next that $X$ contains a maximal linearly independent set $Y$, that is, any other subset of $X$ which contains $Y$ properly is linearly dependent.

Let $\boldsymbol{y}_1$ be a non-zero vector in $X$, and observe that $\left{\boldsymbol{y}_1\right}$ is linearly independent. Now two situations can occur: either (a) every other vector of $X$ lies in the subspace spanned by $\boldsymbol{y}_1$, or (b) we can find $\boldsymbol{y}_2 \in X$ such that $\left{\boldsymbol{y}_1, \boldsymbol{y}_2\right}$ is linearly independent.

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Matrix spaces and spaces of polynomials

$$A=\left[\begin{array}{llll} a_{11} & a_{12} & a_{21} & a_{22} \end{array}\right]$$

$$T: \mathrm{M} 2(\mathbb{K}) \rightarrow \mathbb{K}^4 A \quad \mapsto A \mathcal{B}=\left(a_{11}, a_{12}, a_{21}, a_{22}\right)$$

## 数学代写|线性代数代写linear algebra代考|Existence and construction of bases

Ueft{\boldsymbol{y}_1 1right} 是线性独立的。现在可能会 发生两种情况：要么 (a) $X$ 位于跨越的子空间 $\boldsymbol{y}_1$ ，或者 (b) 我们可以找到 $\boldsymbol{y}_2 \in X$ 这样
Veft{{boldsymbol{y}_1, lboldsymbol{y}__2lright} 是线性独立的。

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