The question I'm working on is:
For the given vectors $v, w$ in $\mathbb R^4$, find other vectors $u_1, …, u_n$ such that {${v, w, u_1,…,u_n}$} form a basis of $\mathbb R^4$.
$v = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$ and $w = \begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix}$
I'm not sure how I'd find other vectors that form a basis for a subspace.
I know the first step is to get the matrix given into Reduced Row Echelon Form, and that all the vectors in a basis are linearly independent. However, putting these two together and reducing leaves:
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0\end{bmatrix}$
Which I'm not sure what to do with to find a basis.
Best Answer
You want the determinant $$ \begin{vmatrix} 1&4&a&w\\ 2&3&b&x\\ 3&2&c&y\\ 4&1&d&z \end{vmatrix}\ne0. $$ Row reduction: If we subtract the previous row from each row, we get $$ \begin{vmatrix} 1&4&a&w\\ 1&-1&b-a&x-w\\ 1&-1&c-b&y-x\\ 1&-1&d-c&z-y \end{vmatrix}. $$ Repeat, and we get $$ \begin{vmatrix} 1&4&a&w\\ 0&-5&b-2a&x-2w\\ 0&0&c-2b&y-2x\\ 0&0&d-2c&z-2y \end{vmatrix} = -5\,\begin{vmatrix} c-2b&y-2x\\ d-2c&z-2y \end{vmatrix}. $$ We need, then, this last $2\times 2$ determinant to be nonzero. If we take $b=x=c=y=0$, we have $$ (c-2b)(z-2y)-(d-2c)(y-2x)=-dy. $$ It is then enough to take $d=y=1$. So, $$ \begin{bmatrix}0\\0\\0\\1\end{bmatrix},\ \ \ \begin{bmatrix}0\\1\\0\\0\end{bmatrix} $$ foot the bill.