Understanding zero vector through an example to find values of t and s

Question

I need to determine if the zero vector is in W. And I'm asked to find values for t & s such that:

My natural intuition (which could be wrong) is to:

• add an additional column to fill out a matrix that is 4 rows
• Then place that rogue 2 of the second row in it's own column
• then set the vector out as:

$\begin{bmatrix} 0 & 1 & -3\\2 & 3 & 0 \\0 & 3 & 1 \\ 0 & 3 & 0 \end{bmatrix} \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}= \begin{bmatrix}0\\0\\0\\0\end{bmatrix}$ and solve for the $\vec{v}$ values like so:

$0v_1+1s-3t=0$

$2v_1 + 3s+0t =0 \rightarrow s=\frac{-2}{3}v_1$

$0v_1+3s-1t=0$

$0v_1+3s + 0t=0$

Such that: $\vec{v} = s\begin{bmatrix} \frac{-2}{3}+1\\3\\3\\3 \end{bmatrix} + t\begin{bmatrix} -3\\0\\1\\0 \end{bmatrix}$

However, it seems I'm worng as the answer sheet says: "The zero vector is not in W. There is no t and s such that the vector equation is satisfied."

Where am I going wrong here, please help me identify my knowledge gap?

Answer 1

The vectors in $W$ can be written as follows

$ w = t \begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 1 \\3\\3\\3\end{bmatrix} + \begin{bmatrix} 0 \\ 2 \\ 0 \\ 0 \end{bmatrix} $

Now we want to find $t , s$ such that $w = 0$, the augmented matrix of this linear system is

$\begin{bmatrix} -3 && 1 && 0 \\ 0 && 3 && -2 \\ 1 && 3 && 0 \\ 0 && 3 && 0 \end{bmatrix}$

By looking the second and fourth rows, we can immediately see that the system is inconsistent (i.e. has no solution), because the second row says

$3 s = -2 $

while the fourth row says

$ 3 s = 0 $

Subtracting, we get $0 \cdot s = -2 $ which has no solution.

Therefore, the zero vector in not in the set $W$, i.e. $0 \notin W$

Answer 2

I don't know why your "natural intuition" is to do things in such a complicated way! You have $\begin{bmatrix}s- 3t \\ 2+ 3s \\ 3s+ t \\ 3s\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ so you must have s- 3t= 0, 2+ 3S= 0, 3s+ t= 0, and 3s= 0. From 3s= 0, s= 0 so the other three equations become s= 0, 2+ 3s= 0, and 3s= 0. There is NO value of s that satisfies all of those.

(I wonder about the "2+ 3s" since that has no "t" in it.)