# Confusion about pivot column basis

Supposed there is a set of vectors $$\{(1,2,3) , (1,0,5), (1,2,1)\},$$ i.e. the matrix$$\begin{pmatrix} 1 & 2&3 \\ 1 & 0&5 \\ 1&2&1 \end{pmatrix},$$ and suppose there is a RREF of these vectors to be $$\begin{pmatrix} 1 & 0&0 \\ 0 & 1&0 \\ 0&0&1 \end{pmatrix}.$$ (I havent worked out the actual RREF)

Then what would the basis be. I know the basis should be pivot columns corresponding to the original matrix but the vectors are arranged in rows.

Note that "the basis should be pivot columns corresponding to the original matrix" is only true if we arrange the vectors by columns and then obtain the RREF matrix by row operations.

In general, when we arrange the vectors in rows, since row operations don't change row space, once we have the RREF matrix, we can choose as basis vectors the pivot rows in the RREF matrix or, as alternative, the corresponding vectors of the original matrix.

In this case, since we have three linearly independent vectors which span $\mathbb{R^3}$, a basis can be made by the original set of vectors $$\{(1,2,3) , (1,0,5), (1,2,1)\}$$

or by the standard one

$$\{(1,0,0) , (0,1,0), (0,0,1)\}$$

• Will adding the fact that the two matrices are equivalent will be a stronger argument then that row operations do not change to row space? – gbox Feb 12 '18 at 14:47
• @gbox what does it mean equivalent matrices? – gimusi Feb 12 '18 at 14:50
• – gbox Feb 12 '18 at 15:34
• @gbox Yes of course by row operation the two matrices are row equivalent in the sense that they share the same rowspace! – gimusi Feb 12 '18 at 15:38