# Homework Help: Basis for matrix

1. Oct 14, 2012

### pyroknife

When you have a matrix like:
3 1
0 1

The RREF is
1 0
0 1
the identity matrix.

Is the basis always [0 0]^t?

2. Oct 14, 2012

### Ray Vickson

No, a basis is never [0 0]^t. A basis always consists of nonzero vectors, and for a 2-D space you need two of them. Anyway, what is a "basis of a matrix"? I have never heard of that term.

RGV

3. Oct 15, 2012

### pyroknife

I was looking at the answers in the back of the book for the attached problem.
The problem statement is "find a basis for the null space of the linear transformation T.

Their answer was {[0 0]^t}. I don't understand how they got this.

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4. Oct 15, 2012

### Ray Vickson

OK, that seems to be a different question from what you first asked. The question is: what is the null space of the matrix? For the given matrix, the null space consists of a single point (x,y) = (0,0); do you know why?

RGV

5. Oct 15, 2012

### pyroknife

Isn't it asking for the basis of the nullspace or is the nullspace of the matrix the same thing?

Hmm, I'm not sure why, it seems if the RREF form has a pivot in every column then this is the case.
Is it cause the augmented matrix would look something like
1 0 0
0 1 0
thus x=0, y=0?

6. Oct 17, 2012

### pyroknife

7. Oct 17, 2012

### Staff: Mentor

The problem is asking for a basis of the nullspace of the transformation. A vector space is different from a basis for that vector space. You can get all of the vectors in some vector space by taking linear combinations of the vectors in a basis.

Here's an example. Suppose that the nullspace happened to be all of the vectors in R2 that lie along the x-axis. A basis for this space is the vector <1, 0>T. This nullspace has infinitely many vectors in it, but all of them are a linear combination (multiples of in the case) the vector <1, 0>T.
Yes.

8. Oct 17, 2012

### HallsofIvy

The question did not ask for a basis of the nullspace, it asked for the nullspace itself. The entire null space consists of the 0 vector. That does not have a basis.