# Maths proof help pls

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Report

#2

This is a "just do it" question - it's hard to give much of a hint.

But for a first step: try to solve (find a matrix that works) for the case where x = (1,0,0,...0); i.e. 1 in the first component, all other components 0.

From there, try to solve for a general x.

But for a first step: try to solve (find a matrix that works) for the case where x = (1,0,0,...0); i.e. 1 in the first component, all other components 0.

From there, try to solve for a general x.

0

reply

Report

#3

(Original post by

Something like this? For the first step?

**ottersandseals1**)Something like this? For the first step?

0

reply

Report

#4

(Original post by

Is this on the right track?

**ottersandseals1**)Is this on the right track?

Also note that what ever you do has to be valid for , but you are using 2x2 matrices so what you're writing would only be valid in .

Last edited by DFranklin; 1 month ago

0

reply

Report

#5

Better, but (a) it doesn't actually work due to algebra errors, (b) what happens if you're dividing by zero, and (c) you will at some point need to generalise to R^n.

0

reply

Report

#6

(Original post by

Aw yeah, if i divide by 0 it will be zero. So it doesn't work. I'm not sure where to go from here

**ottersandseals1**)Aw yeah, if i divide by 0 it will be zero. So it doesn't work. I'm not sure where to go from here

0

reply

Report

#7

(Original post by

Could you give me an idea of how to start it? and then i could try finish it?

**ottersandseals1**)Could you give me an idea of how to start it? and then i could try finish it?

You'll then have to work out how if I told you "only the kth component of my vector is non-xero" you could make a similar solution that only divides by x_k.

0

reply

Report

#8

Getting there: that works if x1 is non-zero. What if x1 is zero but x2 is non-zero? What do you need to change?

Also, don't forget that you are going to need to provide a

Also, don't forget that you are going to need to provide a

**general**solution at some point, so you'll need to think about what you'd do for a general n x n matrix and vector .
0

reply

Report

#9

(Original post by

Wouldn't x2 and x3 always be multiplied by 0 so wouldn't make a difference? I'm stuck with how to turn this to a general solution

**ottersandseals1**)Wouldn't x2 and x3 always be multiplied by 0 so wouldn't make a difference? I'm stuck with how to turn this to a general solution

How would you change it to only depend on x2?

If you really can't work it out, start from a general 3 x 3 matrix A, and assume

**x**= (0, 1, 0), multiply through, and see what you get, and therefore what you'd need to do to force A

**x**=

**y**.

0

reply

Report

#10

(Original post by

i would have a non-trivial column 2?

**ottersandseals1**)i would have a non-trivial column 2?

So you should now be in a position to write your proof. As a hint, it should probably start with a line something like:

Suppose and

**x**is non-zero. Then we can find k such that . Then...

Last edited by DFranklin; 1 month ago

0

reply

Report

#11

(Original post by

so do i write three matrices with non-trivial solutions in different columns?

**ottersandseals1**)so do i write three matrices with non-trivial solutions in different columns?

**all**possible n x n matrices.

It's perfectly fine to say things like "A is the matrix with all columns zero, except..." however.

Last edited by DFranklin; 1 month ago

0

reply

Report

#12

(Original post by

Ive added this. I've still to add the line u mentioned above.

**ottersandseals1**)Ive added this. I've still to add the line u mentioned above.

0

reply

Report

#13

(Original post by

Damn, could you possibly write the solution to the end of it so i can try understand it? It would help immensely

**ottersandseals1**)Damn, could you possibly write the solution to the end of it so i can try understand it? It would help immensely

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top