Creating a basis for U

Question

Consider the subspace

$$ U=\text{span} \lbrace \left[\begin{array}{c} 1\cr 5\cr 0\cr 4 \end{array}\right], \left[\begin{array}{c} 5\cr 1\cr 7\cr -1 \end{array}\right], \left[\begin{array}{c} 3\cr 7\cr 0\cr 12 \end{array}\right] \rbrace $$ of $\mathbb{R}^4$. Create a basis $$ \lbrace \left[\begin{array}{c} -1\cr 3\cr -2\cr 2 \end{array}\right],\left[\begin{array}{c} 2\cr 2\cr 2\cr 2 \end{array}\right],x \rbrace $$ for U. $$ x = \left[\Rule{0pt}{4.8em}{0pt}\right. ? \left]\Rule{0pt}{4.8em}{0pt}\right. $$

I have no clue where to start for this linear algebra problem. I know that the bases for U must be linear independent, however I do not know how to create the last basis of U to be this way. I think it has something to do with the third row since it has 0 zeroes in the spanning set.

Answer 1

Consider the matrix $$ A=\begin{bmatrix} -1 & 2 & 1 & 5 & 3 \\ 3 & 2 & 5 & 1 & 7 \\ -2 & 2 & 0 & 7 & 0 \\ 2 & 2 & 4 & -1 & 12 \end{bmatrix} $$ Perform Gaussian elimination and consider the columns of $A$ corresponding to the pivot columns in the reduced form.

\begin{align} \begin{bmatrix} -1 & 2 & 1 & 5 & 3 \\ 3 & 2 & 5 & 1 & 7 \\ -2 & 2 & 0 & 7 & 0 \\ 2 & 2 & 4 & -1 & 12 \end{bmatrix} &\to \begin{bmatrix} 1 & -2 & -1 & -5 & -3 \\ 0 & 8 & 8 & 16 & 16 \\ 0 & -2 & -2 & -3 & 6 \\ 0 & 6 & 6 & 9 & 18 \end{bmatrix} &&\begin{aligned} R_1&\gets -R_1\\ R_2&\gets R_2-3R_1\\ R_3&\gets R_3+2R_1\\ R_4&\gets R_4-2R_1 \end{aligned} \\[6px] &\to \begin{bmatrix} 1 & -2 & -1 & -5 & -3 \\ 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & -3 & -6 \end{bmatrix} &&\begin{aligned} R_2&\gets \tfrac{1}{8}R_2\\ R_3&\gets R_3+2R_2\\ R_4&\gets R_4-6R_2 \end{aligned} \\[6px] &\to \begin{bmatrix} 1 & -2 & -1 & -5 & -3 \\ 0 & 1 & 1 & 2 & 2 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} &&R_4\gets R_4+3R_3 \end{align} This shows

1. The rank of the matrix is $3$, so indeed the two new vectors you're given belong to $U$
2. The first, second and fourth columns of $A$ are linearly independent.

Thus you can choose $$ x=\begin{bmatrix} 5 \\ 1 \\ 7 \\ -1 \end{bmatrix} $$

Answer 2

We started from $$U=\operatorname{Span}\{u_1,u_2,u_3\}$$

and we want to find $x$ in $\{v_1, v_2, x\}$ such that it is a basis. Note that we can pick $x$ such that $v_1^Tx=0$ and $v_2^Tx=0.$

Since the rank of $U$ is $3$, there exists a non-zero vector $y$ such that $\forall u \in U, u^Ty=0.$

Hence from $v_1^Tx=0, v_2^Tx=0$ and $y^Tx=0$, we should be able to solve for a non-zero $x$.