- #1

- 15

- 0

1 0 2

0 0 7

So I knew that the basis of the image had two dimensions and a null space of one. The first and third columns are linearly independent (or at least thats what I made them)... which means that the second column is a free variable... However, this is where I get confused. The second column is 0, so the basis of the null space is... (lets let the first column be x, the second column be y, and the third column be z) 0y = -1x -2z and 0y = -7z + 0x?

Thus if I were to write it in matrix form it'd be something like:

0

0

0

??? This doesn't seem right though...

I have another question regarding how to find the basis of a subspace. If its written as a matrix I think I can do it, however when its written as an equation I get slightly confused. There are two problems:

A){(x,y,z) are elements of R^3: 7x - 3y + z = 0}

I have to give a basis and the dimension of this equation.

So what I did was:

(7 -3 1) So, z = 3y - 7x. So there are two free variables and one dependent variable, thus the dimension is 2 and the Null space must be 1?

In the same vein of thought:

B){(x,y,z,w) are elements of R^4: x + 2y -z = 5x -4y + w -3z = 0}

I'm not really sure how this equation should be interpreted... Should I just treat it as x + 2y - z = 0, and leave w as a free variable? Thus z = x + 2y, so the dimension is 1 and the Null Space is 3?

I'm not very familiar with this site yet (or linear algebra), so sorry if my formatting is wrong, if you need something clarified or have any questions feel free to ask. Thanks.