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?
Best Answer
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.