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Given the following three vectors:

$$ b_1 = \begin{pmatrix} 1 \\ 3 \\5 \end{pmatrix},\space b_2 = \begin{pmatrix} 2 \\ 1 \\ 7 \end{pmatrix},\space b_3 = \begin{pmatrix} 4 \\ 2 \\3 \end{pmatrix}, \space\space b_1, b_2, b_3 \in \mathbb{R}^3 $$

Does $B = \{ b_1, b_2, b_3 \}$ form a basis of $\mathbb{R}^3$?

I know I can write the vectors as rows of a matrix and then use Gaussian elimination to check for linear independence:

$$ \left[ \begin{array}{ccc|c} 1 & 3 & 5 & 0 \\ 2 & 1 & 7 & 0\\ 4 & 2 & 3 & 0 \\ \end{array} \right] $$

The elimination algorithm itself is clear to me, as is the correct solution. However, I just can't wrap my head around why the vectors are written line-wise into the matrix (and thus form rows). Intuitively, I would've written the vectors as columns of the matrix.

Can anybody give me a brief explanation why they go in line-wise instead of column-wise?

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    $\begingroup$ It gives the same result if you do it columnwise. This is a fundamental observation of linear algebra. There's a few proofs found here. $\endgroup$ Jul 22, 2013 at 19:58
  • $\begingroup$ @ZettaSuro Thanks for the quick reply. If you post your comment as an answer, I can mark it as the correct solution! $\endgroup$ Jul 22, 2013 at 20:01
  • $\begingroup$ No problem. I want to elaborate on it a bit more though, so it might take a minute. $\endgroup$ Jul 22, 2013 at 20:25

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I am using for showing that if the vectors are linearly independent. Here is the way:

$$ \begin{pmatrix} 1 & 3 & 5\\ 2 & 1 & 7\\ 4 &2 &3\\ \end{pmatrix} \underrightarrow{-2R_1+R_2\mapsto R_2}~~\text{&}~~\underrightarrow{-4R_1+R_3\mapsto R_3}\begin{pmatrix} 1 & 3 & 5\\ 0 & -5 & -3\\ 0 &-10 &-12\\ \end{pmatrix}$$

$$ \begin{pmatrix} 1 & 3 & 5\\ 0 & -5 & -3\\ 0 &-10 &-12\\ \end{pmatrix} \underrightarrow{-(1/5)R_2\mapsto R_2}\begin{pmatrix} 1 & 3 & 5\\ 0 & 1 & 3/5\\ 0 &-10 &-12\\ \end{pmatrix}\underrightarrow{\cdots}\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 &0 &1\\ \end{pmatrix}$$ This shows that the space that the vectors in the last matrix produce is also the space that the first vectors produce. But the last matrix has the rank $3$ so it guaranties our vectors are linearly independent on $F=\mathbb R$.

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  • $\begingroup$ Nice work, as usual! + $\endgroup$
    – amWhy
    Jul 23, 2013 at 14:41
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Actually both ways work, but for me column forms more obvious since you just apply the definition in columns if you want check are the vectors independent, you will solve this system :

$$\alpha_{1} v_{1} + \alpha_{2} v_{2}+\alpha_{3} v_{3} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

but it won't show you clearly which vector are dependent.

In rows it will do both things(check if vectors are independent, eliminate dependent vectors) in same time(that is the beauty on it!), because Gaussian elimination will eliminate rows if they are combinations of others. It will leave only independent vectors, so If you have one row consist only 0's you can say the vectors wouldn't be a base. Other wise It would be a base (since dim(B) = 3, and It's independent)

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There is a theorem which says "In an echelon form matrix, no nonzero row is a linear combination of other nonzero rows". So, it shows that what Gauss's linear elimination method eliminates is linear relationships among the rows. If you put vectors as a row vectors in a matrix and then you applies Gauss method, you will eliminate linear relationships among original vectors.

You can find a proof of this in Jim Hefferon's book (http://joshua.smcvt.edu/linearalgebra/, page 53).

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