How do I prove that the linear span of a set of vectors is equal to the set of vectors that satisfy an equation.

Question

The problem states that H is all the vectors in $\mathbb{R}^4$, $[x,y,z,w]^T$, that satisfy the equation: $3x + 4y +z + 2w = 0$

I was given three vectors $\vec{u}=[1, -1, 1, 0]^T, \vec{v}=[0, -1, 2, 1]^T, \vec{w}=[1, -1, -1, 1]^T$.

In the first part of the problem I have proven that the vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ are all in H and that they are linearly independent. For the next part I have to show that span($\vec{u}$, $\vec{v}$, $\vec{w}$) = H.

Is there a way of showing this through calculation?

Right now all I can come up with is that, since $\vec{u}$, $\vec{v}$, and $\vec{w}$ are in H and they are linearly independent, then all possible linear combinations of those vectors would be in H. But there might be other vectors that can satisfy the equation that are not in $\vec{u}$, $\vec{v}$, and $\vec{w}$. Or is there? How can I prove otherwise.

Answer 1

Since $u,v,w$ are linearly independent, you know that $\text{span}(u,v,w)$ has dimension 3.

Since H is the nullspace of the matrix $A=\begin{bmatrix}3&4&1&2\end{bmatrix}$ and $A$ has rank 1, $\dim(H)=\text{nullity}(A)=3$.

Therefore $\text{span}(u,v,w)=H$, since $\text{span}(u,v,w)\subseteq H$ and both subspaces have the same dimension.

Answer 2

H is a proper subspace of $\mathbb R$, as the vector $[1,0,0,0]^T$ doesn't belong to it. So the dimension of H is strictly less than $4$. On the other hand H contains a family of three linearly independent vectors. So the dimension of H is at least equal to $3$.

Finally, the dimension of H is equal to $3$ and H $=\text{ span}(u,v,w)$.