Tiana Hill

2022-09-05

Let ${v}_{1}=\left(\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right)$ , ${v}_{2}=\left(\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right)$, ${v}_{3}=\left(\begin{array}{c}0\\ 0\\ 1\\ 1\end{array}\right)$, ${v}_{4}=\left(\begin{array}{c}2\\ 0\\ 0\\ 1\end{array}\right)$ ${\mathbb{R}}^{4}$ vectors.
Show that every $v\in {\mathbb{R}}^{4×1}$ can be written as vectors $\left({v}_{1},{v}_{2},{v}_{3},{v}_{4}\right)$ linear combination.
My attempt:
$\left[\begin{array}{ccccl}1& 0& 0& 2& {v}_{1}\\ 1& 1& 0& 0& {v}_{2}\\ 0& 1& 1& 0& {v}_{3}\\ 0& 0& 1& 1& {v}_{4}\end{array}\right]$
Where do I go from here? Every input is appreciated.

beshrewd6g

As $detA=-1\ne 0$, given any vector $B=\left({b}_{1},{b}_{2},{b}_{3},{b}_{4}\right)$, we solve the linear system $AX=B$ with
$A=\left(\begin{array}{cccc}1& 0& 0& 2\\ 1& 1& 0& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 1\end{array}\right);\phantom{\rule{thickmathspace}{0ex}}X=\left({x}_{1},{x}_{2},{x}_{3},{x}_{4}\right)$
and thank to Cramer's theorem we get as unique solution the components X of B in the base $\left({v}_{1},{v}_{2},{v}_{3},{v}_{4}\right)$

$|\begin{array}{cccc}1& 0& 0& 2\\ 1& 1& 0& 0\\ 0& 1& 1& 0\\ 0& 0& 1& 1\end{array}|=|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 1& 1\end{array}|-|\begin{array}{ccc}0& 0& 2\\ 1& 1& 0\\ 0& 1& 1\end{array}|=1+|\begin{array}{cc}0& 2\\ 1& 1\end{array}|=-1\ne 0$
so the four vectors are linearly independent, so they span ${\mathbb{R}}^{4}$
so any vector in ${\mathbb{R}}^{4}$ can be expressed as a linear combination of them.

Do you have a similar question?