Originally Posted by

**PiperAlpha167** First, you continue to refer to column vector 1, and ask how __it__ can be linearly independent.

When you say that, what do you have in mind? It doesn't make sense to me.

Certainly, if you throw any nonzero vector into a set by itself, you can say it is a linearly independent set.

(After all, for nonzero vector v, what is the only scalar, c, that will make cv = 0?)

But I don't see what it has to do with this problem.

Second, I claimed that the reduction could be carried out by eye.

Well, I'm sorry to say that I didn't do it myself correctly, so where do I get off suggesting that anyone else should be able to do it correctly?

Certainly, column 2 does not become a column of all zeros, as I claimed.

Trying again (with more care), I see that columns 1, 2, 3, 5 and 6 form a linearly independent set. This is in accord with the matlab result. You can see the unit pivots clearly in those five columns.

So the rank of A is 5. Dimension of column space and row space is each 5; dimension of null space is 2; dimension of left-null space is 4.

Now I expect you will object again, since column 1 is still in this set.

And, you will ask again, how can __it__ be linearly independent.

I think you can answer this question yourself by noting that the column 4 vector is not in the above set.

P.S. I don't think I've made any claim about knowing something (or anything) that you don't already know concerning the notion of linear independence.

Have a good day.