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?
Best Answer
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 $F=\mathbb R$.