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How can I determine if a function lies within the span of a vector valued function?

For an example does $\begin{bmatrix} 2\\ 1\end{bmatrix}$ lie in the span of:

$$a(x) = \begin{bmatrix} 1\\ x\end{bmatrix}$$ $$b(x) = \begin{bmatrix} x\\ 1\end{bmatrix}$$ $$c(x) = \begin{bmatrix} x\\ 2x\end{bmatrix}$$

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2 Answers

up vote 2 down vote accepted

You could approach this problem in an unorganized brute force sort of way like: Suppose $$[2\;1]^T = r[1\;x]^T+s[x\;1]^T+t[x\;2x]^T.$$ This leads to equations: $2=r+sx+tx$ and $1=rx+s+t2x$. Thus $2=r+(s+t)x$ and $1=s+(r+2t)x$. So (equating coefficients) one has $2=r$, $0=s+t$, $1=s$, and $0=r+2t$. Therefore, $r=2$, $s=1$, and $t=-1$. Thus "Yes" $[2\;1]^T=2a(x)+b(x)-c(x)$.

Instead a coordinate approach is a bit more organized (although a little clunkier). Such an approach would be more successful (than brute force) if dealing with a larger problem.

The collection of all functions $\{f:\mathbb{R}\to\mathbb{R}^{2\times 1}\}$ form an infinite dimensional vector space. But your functions lie in a finite dimensional subspace, for example: $W = \left\{ \begin{bmatrix} ax+b\\cx+d \end{bmatrix} \;:\; a,b,c,d \in \mathbb{R} \right\}$. By choosing a basis like $$\beta=\left\{\begin{bmatrix} 1\\0 \end{bmatrix},\begin{bmatrix} x\\0 \end{bmatrix},\begin{bmatrix} 0\\1 \end{bmatrix},\begin{bmatrix} 0\\x \end{bmatrix}\right\}$$ we can write everything in coordinates. So $$ \left[ \begin{bmatrix} 2\\1 \end{bmatrix} \right]_\beta = \begin{bmatrix} 2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \quad \left[ \begin{bmatrix} 1\\x \end{bmatrix} \right]_\beta = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \quad \left[ \begin{bmatrix} x\\1 \end{bmatrix} \right]_\beta = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix} \quad \left[ \begin{bmatrix} x\\2x \end{bmatrix} \right]_\beta = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix}$$. Then asking if $[2\;1]^T$ is in the span of $\{a(x),b(x),c(x)\}$ amounts to asking if the final column of the following matrix lies in the span of the first three columns (i.e. is the final column a non-pivot column?): $$ \begin{bmatrix} 1 & 0 & 0 & : & 2 \\ 0 & 1 & 1 & : & 0 \\ 0 & 1 & 0 & : & 1 \\ 1 & 0 & 2 & : & 0 \end{bmatrix}$$ This matrix row reduces to $$ \begin{bmatrix} 1 & 0 & 0 & : & 2 \\ 0 & 1 & 0 & : & 1 \\ 0 & 0 & 1 & : & -1 \\ 0 & 0 & 0 & : & 0 \end{bmatrix}$$ So the final column is not a pivot column and thus $[1\;2]^T$ lies in the span of $\{a(x),b(x),c(x)\}$. Moreover, the RREF tells us that the desired linear combination is $2a(x)+b(x)-c(x)$ (as before).

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Wow excellent, thank you very much!! –  diimension Sep 19 '12 at 6:37
    
I'm glad it helped. :) –  Bill Cook Sep 19 '12 at 10:55
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Maybe you mean to ask,

Does the point $(2,1)$ fall on the line parametrized by $a(x),b(x)$ or $c(x)$ if any?

$$ a(x) = (2,1)=(1,x) \ \ \Rightarrow \ \ 2=1, 1=x $$ $$ b(x) = (2,1)=(x,1) \ \ \Rightarrow \ \ 2=x, 1=1 $$ $$ c(x) = (2,1)=(x,2x) \ \ \Rightarrow \ \ 2=x, 1=2x $$

Obviously $2=1$ is hard to solve, the equation for $b(x)$ shows $b(2)=(2,1)$ and the equation for $c(x)$ gives $x=2$ and $x=1/2$ which is a contradiction. Only $b$ contains $(2,1)$ in its image.

Following Bill's suggestion: you could find $c_1,c_2,c_3$ such that

$$ (2,1) = c_1(1,x)+c_2(x,1)+c_3(x,2x) $$

This equation has to hold for all $x$ if $(2,1)$ is in the span of $a,b,c$ as vector-valued functions.

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I think he is asking how to tell if the constant function $f(x)=[2\;1]^T$ is in the span of $\{a(x),b(x),c(x)\}$. –  Bill Cook Sep 18 '12 at 1:32
    
well, that would be more interesting. –  James S. Cook Sep 18 '12 at 1:39
    
Lol Bill is right but no worries. Thanks anyways! –  diimension Sep 19 '12 at 6:38
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