Determine whether the vector $(1, 1, 1, 0)^T$ belongs to $U$ or not.

Question

Suppose that the vectors $(1, 2, −1, 0)^T$ and $(1, 3, 2, 0)^T$ span the subspace $U$ of $\mathbb{R}^4$.

Determine whether the vector $(1, 1, 1, 0)T$ belongs to $U$ or not.

Can anyone explain this?

Answer 1

An explanation boils down to $(1,1,1,0)$ not being the zero vector, it isn't a scalar multiple and it isn't the sum of the other two vectors and cannot be obtained from the other two vectors by a combination of scalar multiples and sums. It therefore doesn't meet any of the three criteria for a subspace spanned by the other two vectors.

Answer 2

The vector $x:=\begin{bmatrix}1&1&1&0\end{bmatrix}^T$ belongs to $U:=\operatorname{Span}(\begin{bmatrix}1&2&-1&0\end{bmatrix}^T, \begin{bmatrix}1&3&2&0\end{bmatrix}^T)$ if and only if $x$ is a linear combination of $u_1:=\begin{bmatrix}1&2&-1&0\end{bmatrix}^T$ and $u_2:=\begin{bmatrix}1&3&2&0\end{bmatrix}^T$. Thus, $x\in U$ if and only if there is a solution to $x=au_1+bu_2$ for $a,b\in\Bbb R$, or equivalently, the system of linear equations: $$\begin{cases}1=a+b\\1=2a+3b\\1=-a+2b\\0=0\cdot a+0\cdot b\end{cases}\iff \begin{cases}a+b=1\\2a+3b=1\\-a+2b=1\end{cases}$$