Sum of the components of a vector [duplicate]
Solution 1:
All it means is that you've discovered an invariant property of linear combinations of $\mathbf{v}$ and $\mathbf{w}$, which can then be used to identify whether a vector belongs to that set of combinations.
If you're familiar with the dot product of vectors, then "the components sum to zero" can be interpreted slightly differently - note that $\mathbf{v} \cdot \mathbf{r} = v_x r_x + v_y r_y + v_z r_z = 0$ is satisfied if $r_x = r_y = r_z = 1$, i.e. $\mathbf{r} = (1, 1, 1)$. And if the dot product of two non-zero vectors is zero, then that means that the vectors must be perpendicular to each other.
So the vector $(1,1,1)$ is in fact perpendicular to both $\mathbf{v}$ and $\mathbf{w}$, and as a result it is perpendicular to all vectors of the form $a\mathbf{v} + b\mathbf{w}$, in other words it is perpendicular to the plane formed from all possible linear combinations of $\mathbf{v}$ and $\mathbf{w}$. More generally, for any two vectors $\mathbf{v}$ and $\mathbf{w}$, if the vector $\mathbf{r}$ is perpendicular to both of them then it is perpendicular to all linear combinations of them. This falls out pretty nicely from just noting that $(a \mathbf{v} + b \mathbf{w}) \cdot \mathbf{r} = a \mathbf{v} \cdot \mathbf{r} + b \mathbf{w} \cdot \mathbf{r} = a \times 0 + b \times 0 = 0$
You can even find $\mathbf{r}$ in 3D space pretty quickly, if you're familiar with the cross product, since the cross product of two vectors is always perpendicular to both of them. As it turns out, in this example $\mathbf{v} \times \mathbf{w} = (1, 1, 1)$ which brings us full circle.
Solution 2:
So,
$$\vec{v} = \begin{bmatrix}
1\\
-2\\
1\\
\end{bmatrix}$$
$$\vec{w} = \begin{bmatrix}
0\\
1\\
-1\\
\end{bmatrix}$$
Essentially,
$v_{x}+v_{y}+v_{z} = 0$ and $w_{x}+w_{y}+w_{z} = 0$
Therefore, all elementary row operations will still scale the values and maintain the principle that
$$\sum v_{t}=0$$
where $v$ is a given vector.
$$\vec{\alpha}=
\begin{bmatrix}
cv_{x}+dw_{x}\\
cv_{y}+dw_{y}\\
cv_{z}+dw_{z}\\
\end{bmatrix}
$$
Applying the above postulation:
$$\sum v_{t}=0$$
where $v$ is a given vector,
$$\vec{\alpha}=
\begin{bmatrix}
3\\
3\\
-6\\
\end{bmatrix}
$$
is valid because the sum of components is 0, however,
$$\vec{\alpha}=
\begin{bmatrix}
3\\
3\\
6\\
\end{bmatrix}
$$
Does not satisfy the postulation because the sum of the components is not $0$