# Problem: finding a basis for this vector subspace

My textbook presents the following practice problem:

Let $\vec{v_1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \vec{v_2} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \space$ and $H = \{\begin{bmatrix} s \\ s \\ 0 \end{bmatrix} : s \in \mathbb{R}\}$. Then every vector in $H$ is a linear combination of $v_1$ and $v_2$ because $\begin{bmatrix} s \\ s \\ 0 \end{bmatrix} = s\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. Is the set $\{\vec{v_1}, \vec{v_2}\}$ a basis for $H$?

Our textbook says that a set of vectors $S$ is a basis for a subspace $H$ if:

(i) $S$ is a linearly independent set

(ii) $H = Span(S) = Span\{\vec{s_1}, ..., \vec{s_n}\}$.

I'm struggling to understand why the set $\{\vec{v_1}, \vec{v_2}\}$ is not a basis for $H$. I know the first property is satisfied because the two vectors are in fact linearly independent.

I'm guessing the second property is where this fails, but I don't understand why. If the problem statement itself says that "every vector in $H$" can be written as a linear combination of the two vectors $\vec{v_1}$ and $\vec{v_2}$, then that by definition implies that $H = Span(S)$, doesn't it?

The textbook argues that this is not a basis for $H$ because $\vec{v_1}$ and $\vec{v_2}$ aren't even in $H$, but I don't see how that's relevant to the point we're trying to prove.

Any help is greatly appreciated. Please note that this is my first linear algebra course, so simple answers would be great. Thanks!

• What is the definition of $Span$? – user418131 Oct 7 '17 at 15:08
• It's all possible linear combinations (set) of the vectors $\vec{v_1}$ and $\vec{v_2}$ – AleksandrH Oct 7 '17 at 15:12
• Are all these linear combinations in $H$? – user418131 Oct 7 '17 at 15:13
• I think so, yes. That's where I'm confused. – AleksandrH Oct 7 '17 at 15:15
• Ooooh, I think I've got it now. Thank you! – AleksandrH Oct 7 '17 at 15:38

$\{v_1, v_2\}$ cannot be a basis for $H$ simply because these vector do not belong to $H$.

They cannot be a basis either, because we have an isomorphism $\; K \longrightarrow H$, $\;s \longmapsto (s,s, 0)$ and the cardinal of a basis is an invariant.

• I'm sorry, but this isn't really helpful, as I don't know what cardinal and invariants mean. Also, your first sentence pretty much just restates what the book said, which I already mentioned was unclear to me. Thanks anyway – AleksandrH Oct 7 '17 at 15:47
• The cardinal of a basis is just the number of vectors in a basis, and it's a fundamental result on vector spaces that all bases have the same number of elements. – Bernard Oct 7 '17 at 16:09

If you look at both $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ it is not in $H$: Then YOU CAN NOT GENERATE IT.

$H$ is of dimension $1$ and a base of $H$ forms the vector $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$.

$H$ is a vector line, and two linearly independent vectors generate a vector plane. The line $H$, in this case, is contained in the plane.

In the definition of both generators and base, the vectors are required to be of the set in question.