Maths proof help pls

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#1
This is Q
Last edited by ottersandseals1; 1 month ago
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1 month ago
#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.
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1 month ago
#3
(Original post by ottersandseals1)
Something like this? For the first step?
No. How can this work for an arbitrary vector y? You don't even *mention* y...
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1 month ago
#4
(Original post by ottersandseals1)
Is this on the right track?
Well, kind of but not really. Try the same approach, but with the assumption on x that I gave you. Note that at some point you'll need to actually make some conclusions: you want to be able to say "this is what should equal, this is what should equal, and so on".

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
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1 month ago
#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.
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1 month ago
#6
(Original post by 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
If x isn't zero, there will be *some* component that isn't 0.
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1 month ago
#7
(Original post by ottersandseals1)
Could you give me an idea of how to start it? and then i could try finish it?
If you fix your previous answer, you'll see it crucially depends on a particular component of x not being 0.

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.
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1 month ago
#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 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 .
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1 month ago
#9
(Original post by 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 Ax = y.
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1 month ago
#10
(Original post by ottersandseals1)
i would have a non-trivial column 2?
Yes.

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
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1 month ago
#11
(Original post by ottersandseals1)
so do i write three matrices with non-trivial solutions in different columns?
See above (I edited while you were replying). But, (and this is something I've said about 4 times now): you can't assume you're dealing with 3x3 matrices, so you will have to word a solution that works for 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
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1 month ago
#12
(Original post by ottersandseals1)
Ive added this. I've still to add the line u mentioned above.
I don't understand what you've written. You seem to be dividing by x1,x2,...xn, so now your solution doesn't just depend on x1 being non-xero, but also x2,x3,... In other words, you've made things worse, not better.
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1 month ago
#13
(Original post by ottersandseals1)
Damn, could you possibly write the solution to the end of it so i can try understand it? It would help immensely
Full solutions aren't allowed here. At the end of the day, I give people hints so they can solve a problem themselves. When it becomes me basically answering the question for them, I think they are better off asking their teacher/lecturer.
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