2.4 Coordinates

先给 $V$定义一个ordered basis $\mathscr B=\{\alpha_1,\dots,\alpha_n\}$，其中次序是有用的（即一个sequence），那么对于 $\alpha\in V$，有唯一的一个n-tuple $(x_1,\dots,x_n)$使得 $\alpha=\sum_{i=1}^nx_i\alpha_i$, 于是称 $x_i$是 the $i$th coordinate of $\alpha$ relative to the ordered basis $\mathscr B=\{\alpha_1,\dots,\alpha_n\}$. 总结：事实上 $V$的每个ordered basis 决定了一个一一对应： $\alpha\to(x_1,\dots,x_n)$，并且可以保证加法、数乘也在 $F^n$中对应。
我们将 $\alpha\to(x_1,\dots,x_n)$得到的 $(x_1,\dots,x_n)$记为 $[\alpha]_{\mathfrak B}$, 这个记号在讨论basis的变换时很方便。接下来就进入本节的核心内容：change of basis。如果有 $V$的两组basis： $\mathscr B=\{\alpha_1,\dots,\alpha_n\}, \mathscr B'=\{\alpha_1',\dots,\alpha_n'\}$，每个 $\alpha_j'$都是 $V$中的向量，因此有唯一的scalars $P_{ij}$使得 $\alpha_j'=\sum_{i=1}^nP_{ij}\alpha_i,1\leq j\leq n$，如果我们令 $[\alpha]_{\mathscr B'}=(x_1',\dots,x_n')$，那么可得到
$\alpha=\sum_{j=1}^nx_j'\alpha_j'=\sum_{j=1}^nx_j'\sum_{i=1}^nP_{ij}\alpha_i=\sum_{j=1}^n\sum_{i=1}^n(P_{ij}x_j')\alpha_i=\sum_{i=1}^n\left(\sum_{j=1}^nP_{ij}x_j'\right)\alpha_i$
根据 $[\alpha]_{\mathscr B}$的唯一性，可知 $\sum_{j=1}^nP_{ij}x_j'=x_i$，如果记矩阵 $P=(P_{ij})$，那么 $[\alpha]_{\mathscr B}=P[\alpha]_{\mathscr B'}$或者 $[\alpha]_{\mathscr B'}=P^{-1}[\alpha]_{\mathscr B}$，其中 $P$的可逆性来源于 $[\alpha]_{\mathscr B'}=0\Leftrightarrow [\alpha]_{\mathscr B}=0$，因此根据第一章的Theorem 7， $P$可逆。上述内容则是这一章的Theorem 7的结论，注意到对 $P$有 $P_j=[\alpha_j']_{\mathscr B},j=1,\dots,n$。
以上的讨论从另一个角度，可以得到Theorem 8，即如果先假设 $P$是可逆的，则对于 $V$的一个ordered basis $\mathscr B$, 可以有另一个唯一的ordered basis $\mathscr B'$ 使得 $[\alpha]_{\mathscr B}=P[\alpha]_{\mathscr B'}$或者 $[\alpha]_{\mathscr B'}=P^{-1}[\alpha]_{\mathscr B}$对 $V$中每一个 $\alpha$成立。
Example 18 是standard basis的一个例子。Example 19实际上是二维空间中的rotation变换。从Example 20中可以总结一些规律，如果说矩阵 $P$是可逆的，那么 $P_j=[\alpha_j']_{\mathscr B},j=1,\dots,n$，特别当 ${\mathscr B}$是standard basis时， $P_j$就直接是新的ordered basis ${\mathscr B'}$中的 $\alpha_j'$，求任何一个向量在这个新的basis ${\mathscr B'}$ 中的坐标，就用 $[\alpha]_{\mathscr B'}=P^{-1}[\alpha]_{\mathscr B}$即可，特别是 $X=(x_1,\cdots,x_n)'$在 ${\mathscr B'}$中的坐标是 $P^{-1}X$.

Exercises

1. Show that the vectors

$\alpha_1=(1,1,0,0),\quad \alpha_2=(0,0,1,1) \\\alpha_3=(1,0,0,4),\quad \alpha_4=(0,0,0,2)$

form a basis for $R^4$. Find the coordinates of each of the standard basis vectors in the ordered basis $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$.

Solution: We have
$\begin{aligned}&\begin{bmatrix}1&0&1&0&a\\1&0&0&0&b\\0&1&0&0&c\\0&1&4&2&d\end{bmatrix}\rightarrow\begin{bmatrix}1&0&1&0&a\\0&0&-1&0&b-a\\0&1&0&0&c\\0&0&4&2&d-c\end{bmatrix}\\ \rightarrow &\begin{bmatrix}1&0&1&0&a\\0&0&1&0&a-b\\0&1&0&0&c\\0&0&2&1&\frac{d-c}{2}\end{bmatrix} \rightarrow\begin{bmatrix}1&0&0&0&b\\0&1&0&0&c\\0&0&1&0&a-b\\0&0&0&1&\frac{d-c}{2}+2b-2a\end{bmatrix}\end{aligned}$
Thus the four vectors are linearly independent, since $\dim R^4=4$, we see they form a basis for $R^4$.
by the augmented matrix above, we have
$\begin{aligned}\epsilon_1&=(1,0,0,0)=α_3-2α_4 \\ \epsilon_2&=(0,1,0,0)=α_1-α_3+2α_4 \\ \epsilon_3&=(0,0,1,0)=α_2-1/2 α_4 \\ \epsilon_4&=(0,0,0,1)=1/2 α_4\end{aligned}$

2. Find the coordinate matrix of the vector $(1,0,1)$ in the basis of $C^3$ consisting of the vectors $(2i,1,0),(2,-1,1),(0,1+i,1-i)$, in that order.

Solution:
$\begin{aligned}\begin{bmatrix}2i&2&0&1\\1&-1&1+i&0\\0&1&1-i&1\end{bmatrix}&\rightarrow\begin{bmatrix}1&-1&1+i&0\\0&1&1-i&1\\0&2+2i&2-2i&1 \end{bmatrix}\rightarrow\begin{bmatrix}1&-1&1+i&0\\0&1&1-i&1\\0&0&-2-2i&-1-2i \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&2&1\\0&1&1-i&1\\0&0&1&(3+i)/4\end{bmatrix}\rightarrow\begin{bmatrix}1&0&0&-(1+i)/2\\0&1&0&i/2\\0&0&1&(3+i)/4\end{bmatrix}\end{aligned}$
thus the coordinate matrix is $\begin{bmatrix}-(1+i)/2\\i/2\\(3+i)/4\end{bmatrix}$

3. Let ${\mathscr B}=\{\alpha_1,\alpha_2,\alpha_3\}$ be the ordered basis for $R^3$ consisting of

$\alpha_1=(1,0,-1),\quad \alpha_2=(1,1,1),\quad \alpha_3=(1,0,0)$

What are the coordinates of the vectors $(a,b,c)$ in the ordered basis ${\mathscr B}$?

Solution:
$\begin{aligned}\begin{bmatrix}1&1&1&a\\0&1&0&b\\-1&1&0&c\end{bmatrix}&\rightarrow\begin{bmatrix}1&1&1&a\\0&1&0&b\\0&2&1&a+c\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&1&1&a\\0&1&0&b\\0&0&1&a+c-2b\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&0&b-c\\0&1&0&b\\0&0&1&a+c-2b\end{bmatrix}\end{aligned}$
thus the coordinate matrix is $\begin{bmatrix}b-c\\b\\a+c-2b\end{bmatrix}$.

4. Let $W$ be the subspace of $C^3$ spanned by $\alpha_1=(1,0,i)$ and $\alpha_2=(1+i,1,-1)$

( a ) Show that $\alpha_1$ and $\alpha_2$ form a basis for $W$.

( b ) Show that the vectors $\beta_1=(1,1,0)$ and $\beta_2=(1,i,1+i)$ are in $W$ and from another basis for $W$.

( c ) What are the coordinates of $\alpha_1$ and $\alpha_2$ in the ordered basis $\{\beta_1,\beta_2\}$ for $W$?

Solution:
( a ) It’s obvious that $α_1$ and $α_2$ spans $W$, and they’re linearly independent since they’re not proportionate with each other, thus they form a basis for $W$.
( b ) We have $β_1=-iα_1+α_2$ and $β_2=(2-i) α_1+iα_2$, thus $β_1,β_2\in W$, since they are linearly independent and we already know $\dim W=2$, $β_1,β_2$ form a basis for $W$.
( c ) We have $P=\begin{bmatrix}-i&2-i\\1&i\end{bmatrix}$, and thus $P^{-1}=\frac{1}{2}\begin{bmatrix}1-i&3+i\\1+i&i-1\end{bmatrix}$, let $\mathscr B=\{α_1,α_2\},\mathscr B'=\{β_1,β_2\}$, then
$[α_1]_{\mathscr B}=\begin{bmatrix}1\\0\end{bmatrix},\quad [α_2 ]_{\mathscr B}=\begin{bmatrix}0\\1\end{bmatrix}$
thus
$[α_1]_{\mathscr B'}=\dfrac{1}{2} \begin{bmatrix}1-i&3+i\\1+i&i-1\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}(1-i)/2\\(1+i)/2\end{bmatrix},\\ [α_2]_{\mathscr B' }=\dfrac{1}{2} \begin{bmatrix}1-i&3+i\\1+i&i-1\end{bmatrix} \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}(3+i)/2\\(i-1)/2\end{bmatrix}$

5. Let $\alpha=(x_1,x_2)$ and $\beta=(y_1,y_2)$ be the vectors in $R^2$ such that

$x_1y_1+x_2y_2=0,\quad x_1^2+x_2^2=y_1^2+y_2^2=1$

Prove that ${\mathscr B}=\{\alpha,\beta\}$ is a basis for $R^2$. Find the coordinates of the vectors $(a,b)$ in the ordered basis ${\mathscr B}=\{\alpha,\beta\}$. (The conditions on $\alpha$ and $\beta$ say, geometrically, that $\alpha$ and $\beta$ are perpendicular and each has length 1.)

Solution: To show $\mathscr B=\{α,β\}$ is a basis for $R^2$, it’s enough to show they’re linearly independent, assume they are linearly dependent, then $α=kβ$ or $β=kα$, first let $α=kβ$, then $x_1=ky_1,x_2=ky_2$, thus we have $x_1 y_1+x_2 y_2=ky_1^2+ky_2^2=0$, since $y_1^2+y_2^2=1$, we have $k=0$, so $x_1=x_2=0$, but this contradicts $x_1^2+x_2^2=1$. If we let $β=kα$, we could similarly reach a contradiction.
let $γ=(a,b)$, and let $\mathscr B'=\{\epsilon_1,\epsilon_2 \}$ be the standard basis in $R^2$, then $[γ]_{\mathscr B' }=\begin{bmatrix}a\\b\end{bmatrix}$, and it’s easy to see $α=x_1 \epsilon_1+x_2 \epsilon_2$ and $β=y_1 \epsilon_1+y_2 \epsilon_2$, thus $P=\begin{bmatrix}x_1&y_1\\x_2&y_2 \end{bmatrix}$, and $[γ]_{\mathscr B}=P^{-1} [γ]_{\mathscr B'}$, when $y_2\neq 0$, we shall have
$\begin{bmatrix}x_1&y_1&1&0\\x_2&y_2&0&1\end{bmatrix}\rightarrow\begin{bmatrix}1&0&x_1&x_2\\x_2&y_2&0&1\end{bmatrix}\rightarrow\begin{bmatrix}1&0&x_1&x_2\\0&y_2&-x_1 x_2&x_1^2 \end{bmatrix}\rightarrow\begin{bmatrix}1&0&x_1&x_2\\0&1&\frac{-x_1 x_2}{y_2}&\frac{x_1^2}{y_2}\end{bmatrix}$
If $y_2=0$, then $y_1\neq 0$ and so $x_1=0$, it follows $x_2\neq 0$, in this case we have
$\begin{bmatrix}x_1&y_1&1&0\\x_2&y_2&0&1\end{bmatrix}\rightarrow\begin{bmatrix}x_2&0&0&1\\0&y_1&1&0\end{bmatrix}\rightarrow\begin{bmatrix}1&0&0&1/x_2\\0&1&1/y_1&0\end{bmatrix}$
thus $P^{-1}=\begin{bmatrix}x_1&x_2\\-(x_1 x_2)/y_2 &(x_1^2)/y_2 \end{bmatrix},y_2\neq 0$ and $P^{-1}=\begin{bmatrix}0&1/x_2 \\1/y_1 &0\end{bmatrix},y_2=0=x_1$, and
$[γ]_B=\begin{bmatrix}x_1&x_2\\-(x_1 x_2)/y_2 &(x_1^2)/y_2 \end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}x_1 a+x_2 b\\ \dfrac{x_1}{y_2} (x_1 b-x_2 a)\end{bmatrix},\quad y_2\neq 0 \\ [γ]_B=\begin{bmatrix}0&1/x_2 \\1/y_1 &0\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}b/x_2 \\a/y_1 \end{bmatrix},x_1=y_2=0$

6. Let $V$ be the vector space over the complex numbers of all functions from $R$ into $C$, i.e., the space of all complex-valued functions on the real line. Let $f_1(x)=1,f_2(x)=e^{ix},f_3(x)=e^{-ix}$.

( a ) Prove that $f_1,f_2,f_3$ are linearly independent.

( b ) Let $g_1(x)=1,g_2(x)=\cos x,g_3(x)=\sin x$. Find an invertible $3\times 3$ matrix $P$ such that $g_j=\sum_{i=1}^3P_{ij}f_i$.

Solution:
( a ) Suppose $c_1 f_1+c_2 f_2+c_3 f_3=0$, from $e^{ix}=\cos⁡x+i \sin⁡x$ and $e^{-ix}=\cos⁡x-i\sin⁡x$ we know that
$c_1+c_2 \cos⁡x+c_3 \cos⁡x=0,\quad c_2 \sin⁡x-c_3 \sin⁡x=0$
from the second equation we know $c_2=c_3$, thus $c_1=-2c_2\cos⁡x,\forall x\in R$, if $c_2\neq 0$, then $c_1$ is not constant, a contradiction. Thus $c_2=0$, and it follows $c_1=0,c_3=0$.
( b ) We have
$g_1=f_1,\quad g_2=\dfrac{1}{2} (f_2+f_3 ),\quad g_3=\dfrac{1}{2i}(f_2-f_3)$
thus the invertible matrix $P$ is $P=\begin{bmatrix}1&0&0\\0&\dfrac{1}{2}&\dfrac{1}{2i}\\0&\dfrac{1}{2}&\dfrac{1}{2i} \end{bmatrix}$

7. Let $V$ be the (real) vector space of all polynomial functions from $R$ into $R$ of degree 2 or less, i.e., the space of all functions $f$ of the form $f(x)=c_0+c_1x+c_2x^2$. Let $t$ be a fixed real number and define

$g_1(x)=1,\quad g_2(x)=x+t,\quad g_3(x)=(x+t)^2$

Prove that ${\mathscr B}=\{g_1,g_2,g_3\}$ is a basis for $V$. If

$f(x)=c_0+c_1x+c_2x^2$

what are the coordinates of $f$ in this ordered basis ${\mathscr B}$?

Solution: Suppose $c_1 g_1+c_2 g_2+c_3 g_3=0$, then
$c_1+c_2 (x+t)+c_3 (x+t)^2=0,\quad \forall x\in R \\ c_3 x^2+(2c_3+c_2 )x+(t^2 c_3+tc_2+c_1 )=0,\quad \forall x\in R$
assume $c_1,c_2,c_3$ is not all zero, then the above function is a quadratic function in real coefficients, and it can’t have more than two roots in $R$, a contradiction. Thus we must have $c_1=c_2=c_3=0$, so $\{g_1,g_2,g_3 \}$ is linearly independent.
Let $f=c_0+c_1 x+c_2 x^2$, then we can write
$\begin{aligned}f&=c_2 (x+t)^2+(c_1-2c_2 )(x+t)+c_0-c_1 t+2c_2 t-c_2 t^2\\&=c_2 g_3+(c_1-2c_2 ) g_2+(c_0-c_1 t+2c_2 t-c_2 t^2 ) g_1\end{aligned}$
thus $\{g_1,g_2,g_3\}$ spans $V$, and is a basis of $V$.
From proofs above we can see
$[f]_{\mathscr B}=\begin{bmatrix}c_0-c_1 t+2c_2 t-c_2 t^2\\c_1-2c_2\\c_2 \end{bmatrix}$