How to check if vectors can be part of a basis for a subspace

Question

Question Let V be the subspace of $\mathbb{R}^4$ defined as V={(w, x, y, z) ∈ $\mathbb{R}^4$:w+x+y+z= 0}. Check that the vectors $\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ -1 \\ 0 \\ 1 \end{bmatrix}$ can be part of a basis for V. Then expand the set consisting of these two vectors to a basis of V.

I'm assuming that the answer is that the two vectors can be part of the basis because they are linearly independent, but I'm not sure if I'm correct. Also, how do we know what dimension of V is? Is it correct that dim(V) = 3, since we need a free variable to solve the problem. e.g. w=-x-y-z

Answer 1

Part of basis means not only independent but also spans the space. Dimension of $V =3$ is correct $w=-x-y-z$ written in standard basis as

$x\begin{bmatrix} -1\\0\\0\\0\end{bmatrix}+y\begin{bmatrix}0\\-1\\0\\0\end{bmatrix}+z\begin{bmatrix}0\\0\\-1\\0\end{bmatrix}$

Answer 2

For any vector to be an element of a basis for a vector space $V$, first and foremost it must be an element of that space. Based on your comments you appear to be confusing this will spanning the space. It’s way too early to speak of anything spanning $V$, and this pair of vectors clearly can’t do so, anyway, because per the Rank-Nullity theorem, $V$ is three-dimensional. The two vectors satisfy the defining equation of $V$, so they are indeed elements of $V$.

Second, the vectors must be linearly independent, which should be obvious for these two. Therefore, they can be part of a basis of $V$. As I mentioned above, they cannot span $V$ since you need at least three vectors for that, but we already knew that, anyway, because the next part of the assignment asks to to complete the basis of $V$.

Based on some other comments, it looks like you tried to complete the basis by doing something that you’d no doubt seen for other complete-the-basis: you added the standard basis vectors of $\mathbb R^4$ to the set and then looked for a linearly-independent subset. You’ve overlooked something very important, though: Remember how at the top I said that the first criterion is that the vector be an element of the space? Well, none of the standard basis vectors are elements of $V$, so they’re the wrong thing to use. Instead, you need to supplement the initial set of vectors with other elements of $V$, and then apply the process to that set. An easy-to-find set of vectors in $V$ is $\{(1,-1,0,0)^T, (1,0,-1,0)^T, (1,0,0,-1)^T\}$. The second one of these is obviously a multiple of one of the original two vectors, so we’ll discard that and work with the remaining two. Assembling them into a matrix and row-reducing, we get $$\begin{bmatrix}-1&0&1&1\\0&-1&-1&0\\1&0&0&0\\0&1&0&-1\end{bmatrix} \to \begin{bmatrix}1&0&0&0\\0&1&0&-1\\0&0&1&1\\0&0&0&0\end{bmatrix}$$ so we can use either of the extra vectors to complete the basis.