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10.1: Coordinate Vectors

[ "article:topic", "vettag:vet4", "targettag:lower", "authortag:schilling", "authorname:schilling", "coordinate vector" ]

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Let $$V$$ be a finite-dimensional inner product space with inner product $$\inner{\cdot}{\cdot}$$ and dimension $$\dim(V)=n$$. Then $$V$$ has an orthonormal basis $$e=(e_1,\ldots,e_n)$$, and, according to Theorem9.4.6~\ref{thm:ParsevalsIdentity}, every $$v\in V$$ can be written as

\begin{equation*} v = \sum_{i=1}^n \inner{v}{e_i} e_i. \end{equation*}
This induces a map
\begin{equation*}
\begin{split}
[\,\cdot\,]_e : V &\to \mathbb{F}^n\\
v &\mapsto \begin{bmatrix}
\inner{v}{e_1}\\ \vdots \\ \inner{v}{e_n} \end{bmatrix},
\end{split}
\end{equation*}

which maps the vector $$v\in V$$ to the $$n\times 1$$ column vector of its coordinates with respect to the basis $$e$$. The column vector $$[v]_e$$ is called the coordinate vector of $$v$$ with respect to the basis $$e$$.

Example $$\PageIndex{1}$$:

Recall that the vector space $$\mathbb{R}_1[x]$$ of polynomials over $$\mathbb{R}$$ of degree at most 1 is an inner product space with inner product defined by

\begin{equation*}
\inner{f}{g} = \int_0^1 f(x)g(x)dx.
\end{equation*}
Then $$e=(1,\sqrt{3}(-1+2x))$$ forms an orthonormal basis for $$\mathbb{R}_1[x]$$. The coordinate vector of the polynomial $$p(x)=3x+2\in \mathbb{R}_1[x]$$ is, e.g.,
$[p(x)]_e= \frac{1}{2} \begin{bmatrix} 7 \\ \sqrt{3} \end{bmatrix}.$

Note also that the map $$[\,\cdot\,]_e$$ is an isomorphism (meaning that it is an injective and surjective linear map) and that it is also inner product preserving. Denote the usual inner product on $$\mathbb{F}^n$$ by

\begin{equation*}
\inner{x}{y}_{\mathbb{F}^n} = \sum_{k=1}^n x_k \overline{y}_k.
\end{equation*}

Then

\begin{equation*}
\inner{v}{w}_V = \inner{[v]_e}{[w]_e}_{\mathbb{F}^n}, \qquad \text{for all $$v,w\in V$$,}
\end{equation*}

since

\begin{multline*}
\inner{v}{w}_V = \sum_{i,j=1}^n \inner{\inner{v}{e_i} e_i}{\inner{w}{e_j}e_j}
= \sum_{i,j=1}^n \inner{v}{e_i} \overline{\inner{w}{e_j}} \inner{e_i}{e_j}\\
= \sum_{i,j=1}^n \inner{v}{e_i} \overline{\inner{w}{e_j}} \delta_{ij}
= \sum_{i=1}^n \inner{v}{e_i} \overline{\inner{w}{e_i}} = \inner{[v]_e}{[w]_e}_{\mathbb{F}^n}.
\end{multline*}

It is important to remember that the map $$[\,\cdot\,]_e$$ depends on the choice of basis $$e=(e_1,\ldots,e_n)$$.

Contributors

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