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Let $\alpha_1=[ 2,1,3,0] $ $\alpha_2=[ 1,1,1,-1] $, $\alpha_3=[ 2,-1,5,4] $, $\alpha_4=[ 1,2,0,-3] $, $\alpha_5=[ 3,1,6,1] $ be vectors from $\mathbb{R}^4$ . From vectors system ($\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5 $) choose basis of vector space $V=lin(\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5)\subset\mathbb{R}^4$ spanned by those vectors. I need help with creating a proper matrix for this question.

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  • $\begingroup$ I am trying to understand the question. Are you supposed to choose 4 out of those 5 vectors to create a basis for R4? $\endgroup$ – imranfat Nov 12 '14 at 21:12
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Consider the matrix \begin{bmatrix} 2 & 1 & 2 & 1 & 3\\ 1 & 1 & −1 & 2 & 1\\ 3 & 1 & 5 & 0 & 6\\ 0 & −1 & 4 & −3 & 1 \end{bmatrix} and perform Gaussian elimination on it: \begin{align} \begin{bmatrix} 2 & 1 & 2 & 1 & 3\\ 1 & 1 & −1 & 2 & 1\\ 3 & 1 & 5 & 0 & 6\\ 0 & −1 & 4 & −3 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 1/2 & 1 & 1/2 & 3/2\\ 1 & 1 & −1 & 2 & 1\\ 3 & 1 & 5 & 0 & 6\\ 0 & −1 & 4 & −3 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 1/2 & 1 & 1/2 & 3/2\\ 0 & 1/2 & −2 & 3/2 & -1/2\\ 0 & -1/2 & 2 & -3/2 & 3/2\\ 0 & −1 & 4 & −3 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 1/2 & 1 & 1/2 & 3/2\\ 0 & 1 & −4 & 3 & -1\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{align} The columns where a pivot is found are linearly independent and the others can be written as linear combination of them. So a basis is $$ \{\alpha_1,\alpha_2,\alpha_5\} $$

It's not the only solution, of course, but you can see that $\alpha_5$ will be in any basis extracted from that set, because it is not a linear combination of the other four vectors. If you go on with backwards elimination, \begin{align} \begin{bmatrix} 1 & 1/2 & 1 & 1/2 & 3/2\\ 0 & 1 & −4 & 3 & -1\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} &\to \begin{bmatrix} 1 & 1/2 & 1 & 1/2 & 0\\ 0 & 1 & −4 & 3 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 & 3 & -1 & 0\\ 0 & 1 & −4 & 3 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{align} you see that \begin{align} \alpha_3&=3\alpha_1-4\alpha_2\\ \alpha_4&=-\alpha_1+3\alpha_2 \end{align} because elementary row operation don't change linear relations between columns.

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Since

$$\mathrm{rank}\begin{pmatrix}2 & 1 & 2 & 1 & 3\cr 1 & 1 & −1 & 2 & 1\cr 3 & 1 & 5 & 0 & 6\cr 0 & −1 & 4 & −3 & 1\end{pmatrix}=3$$

the vector subspace $\mathrm{span}\{\alpha_1,\cdots,\alpha_5\}$ has dimension $3.$ So, you need to find three linearly independent vectors.

Now, since

$$\mathrm{det}\begin{pmatrix}2 & 1 & 3\cr −1 & 2 & 1\cr 5 & 0 & 6\end{pmatrix}=24+5-30+6=5\ne 0$$ we have that $\{\alpha_3,\alpha_4,\alpha_5\}$ is a linearly independent set of vectors. So we have got a basis.

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