# Subspaces and the basis for a subspace

$$
\begin{aligned}
\text{Subspace }\mathbf{V} = span&\left(\vec{v\_1}, \vec{v\_2}, \cdots \vec{v\_n} \right)\\
&\begin{Bmatrix} \vec{v\_1}, \vec{v\_2}, \cdots \vec{v\_n} \end{Bmatrix} \text{ is linear independence} \\\\

\text{then}\\\\

\mathbf{S} &= \begin{Bmatrix} \vec{v\_1}, \vec{v\_2}, \cdots \vec{v\_n} \end{Bmatrix} \\
\mathbf{S} &\text{ is a } \bold{Basis}  \text{ for }\mathbf{V}
\end{aligned}
$$

* 若利用 **Minimum set of vectors** 來 span 該 subspace V
* 也就是 span subspace V 的向量都是 linear independence 時
* 這些向量的集合稱為該 Subspace 的 **Basis**

我們舉個例子 T

$$
\mathbf{T} = \begin{Bmatrix} \begin{bmatrix}1\0\end{bmatrix}, \begin{bmatrix}0\1\end{bmatrix} \end{Bmatrix}
$$

首先他可以 span R2 子空間

$$
\begin{aligned}
c\_1\begin{bmatrix} 1\0\end{bmatrix} + c\_2\begin{bmatrix} 0\1\end{bmatrix} &= \begin{bmatrix} x\_1\x\_2\end{bmatrix} \\

c\_1 + 0 = x\_1, c\_1 &= x\_1\\
0 + c\_2 = x\_2, c\_2 &= x\_2\\
\end{aligned}
$$

並且他為 linear independence

$$
\begin{aligned}
c\_1\begin{bmatrix} 1\0\end{bmatrix} + c\_2\begin{bmatrix} 0\1\end{bmatrix} &= \begin{bmatrix} 0\0\end{bmatrix} \\

c\_1 + 0 = 0, c\_1 &= 0\\
0 + c\_2 = 0, c\_2 &= 0\\
\end{aligned}
$$

所以 T 為 R2 的 basis (而且是 ***standard basis***)

而這些 vectors 所生成的任一個向量在 Subspace 中都是獨一無二的 :

$$
\begin{aligned}
\vec{a} &\in \mathbf{V}, \\
\vec{a} &= c\_1\vec{v\_1} + c\_2\vec{v\_2} + \cdots + c\_n\vec{v\_n} \\
\vec{a} &= d\_1\vec{v\_1} + d\_2\vec{v\_2} + \cdots + d\_n\vec{v\_n} \text{ (subtract)} \\

\vec{0} &= (c\_1-d\_1)\vec{v\_1}  + (c\_2-d\_2)\vec{v\_1}  + \cdots + (c\_n-d\_n)\vec{v\_n}
\end{aligned}
$$

因為相減還是在 subspace 裡面，並且滿足 basis (linear independent)，所以 :

$$
\begin{aligned}
c\_1 - d\_1 &= 0 \\
c\_2 - d\_2 &= 0 \\
c\_n - d\_n &= 0 \\
c\_n &= d\_n
\end{aligned}
$$

證明了生成的向量為唯一


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