Nullity of a Linear Map

HellishHeat

Hi, I have an exam in a few hours and I cannot find a good resource for solving this. The example in my notes is very simplistic compared to the sort of questions that have come up in past papers. I would greatly appreciate it if someone could go through this step, by step, as if explaining to a moron

Find the Nullity (Dim of Null space) of the Linear map f:R^4 -> R^3

f(x,y,z,w) = (x-z, y+z-w, 2x+y-z-w)

Then I'll be able to use the Rank+NUllity theorem to get the rank so I'll get at least part of one question correct.

CJC86

Easiest way is to first write the matrix of the linear map, then reduce it to row-echelon form. The number of leading 1s is now the rank of the map and the nullity is going to be (4-rank).

HellishHeat

Thanks for replying but I do not know what you mean. Can you be more specific if you have the time? I don't know how to create that matrix from the linear map and 'm unsure how you can tell the rank by looking at the function. Please elaborate as much as you can afford.

Much obliged.

CJC86

It's quite easy to write the matrix of a linear map. Look at the first component of f, and the numbers in front of x,y,z,w in that order (remembering to include 0s) will be the first row of your matrix. Then second component gives the second row, etc. Then perform Gaussian elimination (basically subtract multiples of rows from each other, one at a time so you don't make a mistake), so that your matrix has only zeroes below the "diagonal". The number of non-zero rows will be your rank, and the nullity is (dimension of source)-(rank).

HellishHeat

Sorry, I'm not getting it. There are no numbers in front of anything. Please illustrate by showing the matrix of the provided Linear Map. I should be able to understand what you mean if I see the matrix.

I can do echelon form and all that follows. I just can't see how one can create a matrix from that Linear map.

CJC86

So, for example, the first row will be (1 0 -1 0).

HellishHeat

you've lost me. Thanks for trying. I've no idea how you got that 1st row.

CJC86

The first component of f(x,y,z,w) is 1.x + 0.y + (-1).z + 0.w, so we get that the first row of the matrix is (1 0 -1 0). The second component is 0.x + 1.y + 1.z + (-1).w and the third component is 2.x + 1.y + (-1).z + (-1).w.

Does that help? We're not meant to give you full answers on here, but let you work it out for yourself.

HellishHeat

I'm not looking for an answer, just a methodology. This is not homework. I have an exam in 2 hours.

I do not know where you have got those 1s and 0s from. They are not mentioned in the problem. That, coupled with the fact that this is not the way it is done in the notes, has further confused me. I think I will just hope the question appears similar to how it has in the notes and I will be able to use the methodology in the notes.

I'm sure your method is sound and I really do appreciate your taking the time to explain it but, perhaps, I'm not sharp enough.

All the best.

CJC86

Sorry if I introduced a new method, but it's the one that I would use myself, and I think it's an easy one to do in all cases once you get it.

Maybe if you post the method in your notes up, and say which parts confuse you?

HellishHeat

Notes:
Find the rank and nullity of the linear map f : R4 −! R3
f(w, x, y, z) = (w + x, x + y, y + z).
First we calculate the nullity.
ker f is the solution set of
w + x = 0
x + y = 0
y + z = 0.
Easy to see that w = y = −x = −z, i.e.,
ker f = {(,−, ,−) : 2 R}
ker f = {(1,−1, 1,−1) : 2 R}

[end]

I think I've just copped on to what you are doing though. It does seem better.

Thanks again

CJC86

Ew, I can see why you'd be confused by that. Basically at that point, you see that all the other variables only depend on what w is, so w is free to be any real number t, but then we must have x=-t, y=t, z=-t, so the dimension of the null space is 1. If we get that the solution space allows 2 free variables, then the nullity is 2, etc.

If you get the method that I was suggesting, then I would encourage you to use it, as it always works out pretty quickly. Otherwise I hope that I have at least helped to explain what is happening in your notes. Good luck in the exam.