# 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.

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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