Find a basis that is not a subset of the spanning set

Question

Consider the subspace of $\mathbb{R}^4$ given by $W$=span$\big\{\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}3\\2\\1\\1\end{bmatrix},\begin{bmatrix}1\\2\\3\\1\end{bmatrix},\begin{bmatrix}0\\2\\4\\1\end{bmatrix} \big\}$.

The first part of the question asks me to find a subset $S$ of the spanning set that is a basis for $W$. I did that by combining the vectors by column and row reducing them, and got $\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}3\\-2\\0\\0\end{bmatrix},\begin{bmatrix}1\\2\\-2\\0\end{bmatrix},\begin{bmatrix}0\\4\\-3\\0\end{bmatrix}$. I got this method from this thread. Since $v_4=1.5v_3-0.5v_2$, this set is linearly dependent, so $v_1,v_2,v_3$ span. However, I'm unsure because the last entry of the vectors are all $0$, which cannot be a basis for $W$ since a vector spanned by the spanning set of $W$ can be $2$, for example, if we add $v_1+v_2$ and get $2$ in the last entry.

The next part asks me to find a basis that is not a subset. I think I just need to find a basis such that the last row is $0$ and not a subset of $W$, but I'm not sure. If my logic is correct, then I want to find four linearly independent vectors such that the last entry of all four vectors is $0$. I cannot see how I can find one without being linearly dependent.

Answer 1

Looking at the thread you referenced, and using your work so far, since columns $1,2$ and $3$ are a basis after row reduction, you can conclude that the first three vectors (from the original set), for instance, are a basis that fulfills your requirements... note that vectors $1,2$ and $4$ would also work (so would $1,3$ and $4$, or $2,3$ and $4$, it appears) ...

Then, for the second part, just change one or more vectors by multiplication by a scalar other than $0$ or $1$... (this will preserve linear independence).

Answer 2

What was the subset you found for the first part of the question?

Your solution is not correct, since the first and last vectors are the same. Hence, they are not linearly independent and thus cannot form a basis. However, by definition, a basis is a set of vectors that is linearly independent and spans the space. Hence, you need to find $d$ linearly independent vectors where $d$ is the dimension of $W$ which you should have found in your first part. One way to do that is to start from the basis that you found in the first part and then exchange vectors in your basis with vectors from the space $W$ while preserving linear independence.

Answer 3

I haven't done the calculations for your specific question, but say for example $W$ works out to be $3$-dimensional and a basis is $\{u,v,w\}$, a subset of the given set. Then for example $\{2u,3v,4w\}$ is also a basis which is not a subset of the given set.

Your proposed answer cannot be right as it is linearly dependent.

Just a thought... are you sure the question wasn't "find a subset which is not a basis"?