# Finding basis using 4 vectors with 3 columns, 1 row??

This is what I'm having a little bit of trouble with this question: given four

\begin{bmatrix}{3}\\{2}\\{-4}\end{bmatrix}\begin{bmatrix}{-6}\\{1}\\{7}\end{bmatrix}\begin{bmatrix}{0}\\{-5}\\{2}\end{bmatrix}\begin{bmatrix}{3}\\{7}\\{-5}\end{bmatrix}

Now the question asks for a basis of the subspace spanned by the {u,v,w,z}, which are the respective vectors I posted above.

The book places the above vectors into ROWS, not columns and I am a bit perplexed why. The correct answer here is just the first 3 vectors, but IM curious why the entires are placed in columns, not rows. The row reduction math is easy.

This is the form they started the row operations:

\begin{bmatrix}{3}&{2}&{-4}\\{-6}&{1}&{7}\\{0}&{-5}&{2}\\{3}&{7}&{-5}\end{bmatrix}

• Better to put them in columns and perform RREF. Three is a party, four is a crowd. One needs to go. I also don't know why they are put in columns. And I can see that the vectors aren't happy about it...:) – imranfat Apr 23 '16 at 0:37
• Well you cant just eliminate one, so that's why they entered them this way . But is there any particular logic behind it? – Charles Johnson Apr 23 '16 at 0:51
• I don't see the logic to put them in that notation. Like you said, put them behind each other and perform RREF. – imranfat Apr 23 '16 at 0:53

## 1 Answer

Elementary row operations (EROs) preserve the row space of a matrix. So sticking the vectors in the rows of a matrix and performing EROs will be a very convenient way to get a basis for the space. Not only will you get a basis, but you'll often get a nicer basis (more zeros) than just figuring out a maximal linearly independent subset of your original vectors.

• Some textbooks consistently put vectors in rows, others use columns. Strictly speaking for its purpose, it doesn't really matter for the question. The same I see when dealing with Markov and Leontief matrices. Column notation works nice for the TI, so I always put vectors in columns. . – imranfat Apr 23 '16 at 1:11