I don't understand the answer to the exercise that I'm doing, can anyone explain it in a more clear manner?
Let $ v_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, v_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, v_3 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $, and let $H$ be the set of vectors in $\Bbb R^3$
whose second and third entries are equal. Then every vector in $H$ has
a unique expansion as a linear combination of $v_1, v_2, v_3$, because$ \begin{bmatrix} s \\ t \\ t \end{bmatrix} = s \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + (t-s) \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + s \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $
for any $s$ and $t$. Is $\{v_1,v_2,v_3\}$ a basis for $H$?Why or why not?
The answer given was:
In order for the set to be a basis for $H$, $\{v_1,v_2,v_3\}$ must be a spanning set for $H$; that is, $H=\operatorname{Span}\{v_1,v_2,v_3\}$. The exercise shows that $H$ is a subset of $\operatorname{Span}\{v_1,v_2,v_3\}$, but there are vectors in $\operatorname{Span}\{v_1,v_2,v_3\}$ which are not in $H$ ($v_1$ and $v_3$, for example). So $H \neq \operatorname{Span}\{v_1,v_2,v_3\}$, and $\{v_1,v_2,v_3\}$ is not a basis for $H$.
I see $v_1$ and $v_3$ making $H$, why does it say they are not in $H$?
Best Answer
You'll notice that $H$ is indeed a subset of $\Bbb R^3$, but to be more precise, it's a subspace of it resembling $\Bbb R^2$ (it's a plane). To see this, notice that the first variable is free, and the third dependent upon the third.
Now, you can use the vectors $v_1, v_2, v_3$ to express points in that plane, but you'll notice that you can also use those vectors to express points that are not in the plane, $v_1$ for example, so $Span(v_1,v_2,v_3)$ does not span $H$.