MATH 240

Mon. October 21st, 2019

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Unique Representation Theorem

Theorem. Let $B=\{b_1,...,b_n\}$ be a basis for a vector space $V$. Then for each $x$ in $V$, there exists unique $c_1,...,c_n$ such that $x=c_1b_1+...+c_nb_n$.

Coordinates

Definition. Suppose $B=\{b_1,...,b_n\}$ is a basis for $V$ and $x$ is in $V$.
The coordinates of $x$ relative to the basis $B$ (or the $B$-coordinates of $x$) are the weights $c_1,...,c_n$ such that $c_1b_1+...+c_nb_n=x$.
$[x]_B=\begin{pmatrix} c_1\\...\\c_n \end{pmatrix}$

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From $B$-Coordinates to Standard Basis Coordinates

Ex. If $b_1=\begin{pmatrix} 1\\0 \end{pmatrix},\ b_2=\begin{pmatrix} 1\\2 \end{pmatrix},\ B=\{b_1,b_2\},$ and $[x]_B=\begin{pmatrix} -2\\3 \end{pmatrix}$, then find $x$. $\begin{aligned} x&=c_1b_1+c_2b_2\\ &=-2\begin{pmatrix} 1\\0 \end{pmatrix}+3\begin{pmatrix} 1\\2 \end{pmatrix}\\ &=\begin{pmatrix} 1\\6 \end{pmatrix}. \end{aligned}$

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From Standard Basis Coordinates to $B$-Coordinates

Ex. If $b_1=\begin{pmatrix} 2\\1 \end{pmatrix},\ b_2=\begin{pmatrix} -1\\1 \end{pmatrix},\ B=\{b_1,b_2\},$ and $x=\begin{pmatrix} 4\\5 \end{pmatrix}$, then find $[x]_B$. $\begin{aligned} x&=c_1b_1+c_2b_2\\ \begin{pmatrix} 4\\5 \end{pmatrix}&=c_1\begin{pmatrix} 2\\1 \end{pmatrix}+c_2\begin{pmatrix} -1\\1 \end{pmatrix}\\ \begin{pmatrix} 4\\5 \end{pmatrix}&=\begin{pmatrix} 2&-1\\1&1 \end{pmatrix}\begin{pmatrix} c_1\\c_2 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 2&-1&4\\1&1&5 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&1&5\\2&-1&4 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&1&5\\0&-3&-6 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&1&5\\0&1&2 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&0&3\\0&1&2 \end{pmatrix}\\ [x]_B&=\begin{pmatrix} c_1\\c_2 \end{pmatrix}\\ &=\begin{pmatrix} 3\\2 \end{pmatrix}. \end{aligned}$

Theorem. Let $P_B=\begin{bmatrix} b_1&...&b_n \end{bmatrix}$. Now $x=P_B[x]_B$.
The columns of $P_B$ are linearly independent since they form a basis, so by the Invertible Matrix Theorem, $P_B$ is invertible. This means that $[x]_B=P_B^{-1}x$.

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Isomorphism

Let $B=\{b_1,...,b_n\}$ be a basis for $V$. Assigning coordinates to $V$ gives a linear transformation from $V$ to $\R^n$ that is one-to-one and onto. This means that $V$ and $\R^n$ are isomorphic – they look different, but can be represented by the same mathematical object.

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Ex. Let $P_5$ be the set of all polynomials of degree five or less.
$\{1,t,t^2,t^3,t^4,t^5\}$ is a basis for $P_5$. This means any polynomial of degree five or less $P$ takes the form $P(x)=a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x^1+a_0$.

If $T$ is the transformation that assigns coordinates from $P_5$ to $\R^6$, then $T(P(x))=\begin{pmatrix} a_5\\a_4\\a_3\\a_2\\a_1\\a_0 \end{pmatrix}$
Thus $P_5$ is isomorphic to $\R^6$. This means we can perform computations like adding two polynomials in $P_5$ by adding their corresponding vectors in $\R^6$.

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Ex. $B=\{\sin{t},\ \cos{t}\}$ is the basis of a subspace inside of $C^\infin(\R)$. ($\sin{t}$ and $\cos{t}$ are linearly independent functions.) Since $B$ only contains two vectors, functions within this subspace can be represented as vectors in $\R^2$.