$\newcommand{\Span}{\operatorname{Span}}$Let $V$ be a vector such that $\dim V = n$, and let $v_1,\ldots,v_k \in V$ be independent vectors such that $1<k\leq n$. Now Let $w_1,\ldots,w_r\in$ $\Span\left\{ v_1,\ldots,v_k \right\}$ be independent vectors such that $1\leq r <k$.
My question is this: How do I find the missing vectors $w_{r+1},\ldots,w_k\in \Span\left\{ v_1,\ldots,v_k \right\}$ so that $\Span\left\{w_1,\ldots,w_r,w_{r+1},\ldots,w_k\right\} = \Span\left\{ v_1,\ldots,v_k \right\}$ ?
Now I know they exist, I just don't know how I actually find them.
For example: Lets look at $\mathbb{R}^5$ and $$U=\Span\left\{ \begin{pmatrix} 5\\2\\3\\7\\3 \end{pmatrix},\begin{pmatrix} 2\\4\\4\\8\\1 \end{pmatrix} ,\begin{pmatrix} 3\\4\\7\\6\\1 \end{pmatrix},\begin{pmatrix} 5\\8\\6\\4\\8 \end{pmatrix} \right\}$$
Those are all independent vectors. Now lets take $2$ independent vectors that are linear combination of those. lets say, $$w_1 = \begin{pmatrix} 6\\2\\11\\15\\-3 \end{pmatrix} \; w_2=\begin{pmatrix} 0\\4\\-2\\12\\1 \end{pmatrix}$$ How can we complete those two vectors to form a basis of $U$ ?
I would really like to understand the general idea of this. This is actually a general question of something I need it for, which is the process of finding a Jordan basis for matrices\transformations.
Thanks for any help
Best Answer
Here's one method for the case of $V = \Bbb R^n$: row-reduce the matrix $$ \pmatrix{w_1 & \cdots & w_r & v_1 & \cdots & v_k} $$ The columns of this matrix that eventually become pivot columns are precisely those which are linearly independent to the preceding columns. It suffices, then, to take these columns to form your basis.