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Linear combinations of four-dimensional vectors

Mathematics Asked by Sammy on December 18, 2020

Let $v_1 = begin{pmatrix} 1 \ 1 \ 0 \ 0 end{pmatrix}$ $v_2 = begin{pmatrix} 0 \ 1 \ 1 \ 0 end{pmatrix}$ $v_3 = begin{pmatrix} 0 \ 0 \ 1 \ 1 end{pmatrix}$ $v_4 = begin{pmatrix} 2 \ 0 \ 0 \ 1 end{pmatrix}$ $mathbb{R}^4$ vectors.

Show that every $v in mathbb{R}^{4times1}$ can be written as vectors $(v_1,v_2,v_3,v_4)$ linear combination.

My attempt:

$left[begin{array}{cccc|l}
1 & 0 & 0 & 2 &v_1\
1 & 1 & 0 & 0 & v_2\
0 & 1 & 1 & 0 &v_3\
0 & 0 & 1 & 1 &v_4\
end{array}
right]$

Where do I go from here? Every input is appreciated.

2 Answers

As $det A=-1ne 0$, given any vector $B=(b_1,b_2,b_3,b_4)$, we solve the linear system $AX=B$ with $$A=left( begin{array}{cccc} 1 & 0 & 0 & 2 \ 1 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 1 \ end{array} right);;X=left( x_1,x_2,x_3,x_4 right)$$ and thank to Cramer's theorem we get as unique solution the components $X$ of $B$ in the base $left(v_1,v_2,v_3,v_4right)$.

Correct answer by Raffaele on December 18, 2020

$begin{vmatrix}1&0&0&2\1&1&0&0\0&1&1&0\0&0&1&1end{vmatrix}=begin{vmatrix}1&0&0\1&1&0\0&1&1end{vmatrix}-begin{vmatrix}0&0&2\1&1&0\0&1&1end{vmatrix}=1+begin{vmatrix}0&2\1&1end{vmatrix}=-1ne0$,

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.

Answered by J. W. Tanner on December 18, 2020

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