$U = \{(x, y, z, w)\in R^4 | x – 2y = z+w\}$
$S = \{(1, 0, 1, 0), (2, 1, 0, 0), (1, 0, 0, 1)\}$
Determine if S is a basis for U.
I started by showing that S is linearly independent. I made a matrix where each vector in S was a column. I found that the only solutions for $a(1, 0, 1, 0) + b(2, 1, 0, 0) + c(1, 0, 0, 1) = 0$ were $a, b, c = 0$.
Now that I know S is linearly independent, how do I show that it is or isn't a basis of U? I checked and know that every vector in S is in U.
Best Answer
For $S$ to be a basis, it must be linearly independent and span $U$.
As you noted, linear independence can be checked by computing the kernel of the matrix.
To check if $S$ spans $U$, let $\vec{u}$ be an arbitrary vector in $U$ and show that $\vec{u}\in\text{span}(S)$.
To this end, a general vector of $U$ is of the form: $$\begin{bmatrix}2y+z+w \\ y \\ z \\ w\end{bmatrix}.$$ So we want to find $a,b,c\in\mathbb{R}$ such that $$\begin{bmatrix}2y+z+w \\ y \\ z \\ w\end{bmatrix}=a\begin{bmatrix}1 \\ 0 \\ 1 \\ 0\end{bmatrix}+b\begin{bmatrix}2 \\ 1 \\ 0 \\ 0\end{bmatrix}+c\begin{bmatrix}1 \\ 0 \\ 0 \\ 1\end{bmatrix}$$
To do this, try and check which vectors on the right-hand side contain nonzero entries which are zero for the other two vectors. For example, the first vector contains a $1$ in the third row, whereas all the other vectors have zero in the third row! Because of this, we know that $a$ equals $z$.
Keep doing this process. Eventually, you'll either find $a,b,c$ that satisfy this, or you'll be able to make up a counterexample! (In this case such $a,b,c$ making this equation hold exist, so there is no counterexample!)
Then since $S$ is linearly independent and spans $U$, $S$ is a basis for $U$!