Another problem I'm trying to get. Here's what it says.
Let $W$ be the set of all vectors of the form shown, where $a$, $b$, and $c$ represent arbitrary real numbers. In each case, either find a set $S$ of vectors that spans $W$ or give an example to show that$W$ is not a vector space.
(a) $$\begin{bmatrix}−a+1\\ a−6b\\ 2b+a\\ \end{bmatrix}$$
(b) $$\begin{bmatrix}4a+3b\\ 0\\ a+b+c\\ c−2a\\ \end{bmatrix}$$
I tried to solve (a) (and say that W is not in the vector space because of the zero vector rule) by doing the following
$-a+1 = 0$
$-a=-1$
$a=1$
Then I used a=1 to substitute into the next part.
$a-6b=0$
$1-6b-0$
$-6b=-1$
$b=1/6$
Then I substituted a=1 and b=1/6 into the next part.
$2b+a=0$
$2(1/6)+1=0$
$(1/3)+(3/3)=0$
$(4/3) \ne 0$
Obviously, I can't use this method for (b) as easily. Is there another way to do this? Am I on the right track with (a)?
(a) looks right, hint for (b): $\quad\quad \begin{bmatrix}4a+3b\\ 0\\ a+b+c\\ c−2a\\ \end{bmatrix} = a \cdot \begin{bmatrix}4\\ 0\\ 1\\ −2\\ \end{bmatrix} + b \cdot \begin{bmatrix}3\\ 0\\ 1\\ 0\\ \end{bmatrix} + c \cdot \begin{bmatrix}0\\ 0\\ 1\\ 1\\ \end{bmatrix} $