# Wolfram alpha wrong on linear independence for 4d vectors?

The first three vectors in the following statement are linearly independent. I put this statement into Wolfram Alpha and it tells the 4 vector set is linearly independent

linear independence of {(1,2,0,2), (1,1,1,0), (2,0,1,3),(1,1,1,1)}

However when I used them as columns of a matrix and row reduced it I found they were linearly dependent.

I then used a completely random number for the 4th vector -

linear independence of {(1,2,0,2), (1,1,1,0), (2,0,1,3),(1,20,156,133)}

and it tells me this is linearly independent too. So is Wolfram Alpha wrong?

-
A "completely random" fourth vector added to a set of three linearly independent vectors would almost always form a linearly independent set - so I'm not surprised by WA's result. – Ted Apr 20 '12 at 8:05
Inputting MatrixRank[{{1, 2, 0, 2}, {1, 1, 1, 0}, {2, 0, 1, 3}, {1, 1, 1, 1}}] into Alpha confirms that the four vectors are independent. Can you show what you did in your row reduction? – J. M. Apr 20 '12 at 8:06
@J.M. I have found a mistake I made during row reduction. Hmmm, I think I've been confusing linear independence with orthogonality. Is it correct to say that if a vector is orthogonal to other vectors the set is linearly independent, however if a vectors is linearly independent it doesn't necessarily mean that it's orthogonal to the other vectors? – Jim_CS Apr 20 '12 at 8:14
Try using RowReduce[{{1, 2, 0, 2}, {1, 1, 1, 0}, {2, 0, 1, 3}, {1, 1, 1, 1}}] – Scott Carter Apr 20 '12 at 9:09
@Jim_CS Yes, orthogonality is a stronger condition than linear independence. (orthogonal implies linear independence but converse is false) – yoyostein Apr 20 '12 at 9:37