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I am reviewing for my linear algebra final and I completely forgot how to do these types of problems. How can I determine if a function lies within the span of a vector valued function?

For example, does $\begin{bmatrix} 1-2x\\-1-x \end{bmatrix}$, $\begin{bmatrix} 1-2x\\1-x \end{bmatrix}$,$\begin{bmatrix} 2-x\\0\end{bmatrix}$ lie in the span of:

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

By using Gaussian elimination?

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up vote 3 down vote accepted

There is no difference in method as opposed to vectors with constant entries, except that it may be necessary to work with algebraic expressions and that sometimes it may be necessary to distinguish cases (if an expression is zero or not).

But some tricks are possible: If you check e.g. the case $x=0$, you may just "see" an easy way to combine (and also see what degree of freedom is left and must be suitably chosen to work for all $x$).

Thus for the first vector with $x=0$, we find $$\left[{1\atop -1}\right]=1\cdot a(0)+(-1)\cdot b(0)+t\cdot c(0)$$ with + for the moment - arbitrary $t$ and notice that $$1\cdot a(x)+(-1)\cdot b(x)+t\cdot c(x)=\left[{1-x+tx\atop -1+x+2tx}\right]$$ so obviously we need both $t-1=-2$ and $2t+1=-1$, which has the valid (and unique) solution $t=-1$. Thus $$1\cdot a(x)+(-1)\cdot b(x)+(-1)\cdot c(x)=\left[{1-2x\atop -1-x}\right].$$

So one noticeable difference in the end is that the three coefficients are uniquely determined although the vectors have only two rows. This is no contradiction because the space of 2D vector functions with affine linear entries $\left[{\alpha+\beta x\atop\gamma+\delta x}\right]$ is in fact four-dimensional.

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Thank you very much, but I am a bit confused. What method did you use to solve this? There isn't any coherent, organized way of solving this, like putting it in matrix form and using gaussian? Also, I am lost on how you got $(-1) \dot\ b(0)$ and $t \dot\ c(0)$? – diimension Dec 6 '12 at 7:37
Yes,admittedly, I used the method of staring at the numbers. But Gauss is of course a formalized ways of doing this. In that case, I'd suggest to map $\left[{\alpha+\beta x\atop\gamma+\delta x}\right]\mapsto (\alpha,\beta,\gamma,\delta)$ and then solve the linear dependencies in plain old $\mathbb R^4$: Is $(1, -2, -1,-2)$ in the span of $\{(1,0,0,1),(0,1,1,0),(0,1,0,2)\}$? – Hagen von Eitzen Dec 6 '12 at 19:17
Sry to keep bothering on this question but why does it span $\mathbb{R^4}$ instead of $\mathbb{R^2}$? How did you get those basis? – diimension Dec 6 '12 at 21:58

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