Thread: Null space and linear independence

My linear algebra textbook claims that if the zero-vector is the only element in the null-space of a matrix, the following has to be true:

1) The column vectors of the matrix are linearly independent.
2) The reduced row echelon form of the coefficient matrix has to be an nxn identity matrix.

Point 1 is obvious to me, but I do not get why the matrix has to be a square matrix. If I have a, say, 3x2 matrix, and it's reduced row echelon form is:

10
01
00

Won't the column vectors in this case be linearly indepentent, and the only solution to the equation Ax=0 be the 2x1 zero-vector? Would appreciate any clarification!

delete

It follows from the rank nullity theorem.
Rank
rank A + nullity A = n
Where the nullity of A is the dimension of the null space of A.

Thanks, but I couldn't make much sense of your link, as I've only read half of my textbook so far. Did you mean that the nullspace for the transpose of A will be the same as the nullspace of A?

Well, what I was meaning is that if you count the number of independent rows of a matrix and then add the dimension of the null space you should get , the number of rows. So if the null space only has the zero vector it has dimension zero. Therefore the rank of the matrix has to be equal to the number of rows. On the other hand we know that the Row Rank is always equal to the Column Rank. Since the null space has only the zero vector in it, we also know that all the columns are independent. If all the columns are independent we know that the Column Rank is equal to the number of columns. Thus the number of columns must be equal to the number of rows.

I hope that wasn't confusing.

And no the transpose of A does not have to have the same null space as A. However, if A is invertible then the transpose will invertible and therefore the null space will only contain the zero vector. So clearly in that case the null spaces are equal.