# Hard(ish) Matrix questionWatch

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#1

I can sort of do it by trial and error but I'm struggling to find a better and more systematic approach.

All I've done is write the Matrix A in terms of a, b, c, d and write out 4 expressions equal to zero.
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#2
bump
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1 year ago
#3
If A^2 = 0 then the eigen-values of A must all be 0 (aka characteristic poly x^2 = 0). Maybe this will help you.
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1 year ago
#4
(Original post by RuneFreeze)

I can sort of do it by trial and error but I'm struggling to find a better and more systematic approach.

All I've done is write the Matrix A in terms of a, b, c, d and write out 4 expressions equal to zero.
This is a bit of a silly question really isn't it? Why would you do part (a) if you can just do part (b) and then repeat your answer :P.

Anyway, I am not sure there is a particularly better way of doing it. Let your matrix be
.
Now because we have some choice of scale here and all the entries are non-zero, we can choose . Then if you calculate you get some simultaneous equations.
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#5
(Original post by Louisb19)
If A^2 = 0 then the eigen-values of A must all be 0 (aka characteristic poly x^2 = 0). Maybe this will help you.
Hmm I haven't (or won't) do eigenvalues. This is only the second chapter I've done on matrices (matrix multiplication) so it should just require very basic matrices stuff.
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1 year ago
#6
name the entries a,b,c,d then eigenvals satisfy

x^2 - x(a+d) + (ad - bc) = 0

so a = -d and then choose b,c such that (ad - bc) = 0. Then you are all good since then eigvals are both 0.
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1 year ago
#7
(Original post by RuneFreeze)
Hmm I haven't (or won't) do eigenvalues. This is only the second chapter I've done on matrices (matrix multiplication) so it should just require very basic matrices stuff.
oh nevermind then.
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#8
(Original post by Deranging)
This is a bit of a silly question really isn't it? Why would you do part (a) if you can just do part (b) and then repeat your answer :P.

Anyway, I am not sure there is a particularly better way of doing it. Let your matrix be
.
Now because we have some choice of scale here and all the entries are non-zero, we can choose . Then if you calculate you get some simultaneous equations.
Ok so I've just done four simultaneous equations by multiplying

which give the simultaneous equations

From the second and third equation, when

So we know that the solution must follow from
and

So I have reduced the four simultaneous equations to two simpler ones, at this point I guess the only method is trial and error which is simple enough.

e.g.

Excuse the superfluousness of this btw, just wanted to try and learn latex

Thanks everyone for responses
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1 year ago
#9
(Original post by RuneFreeze)
From the second and third equation, when
Your working is fine, but this is not a correct statement. Just because it does not imply that .
You might just be misusing the symbol here, but this is just a small pick on your working.
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1 year ago
#10
(Original post by RDKGames)
Your working is fine, but this is not a correct statement. Just because it does not imply that .
You might just be misusing the symbol here, but this is just a small pick on your working.
From the context, I suspect it was just a typo, and they meant
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1 year ago
#11
(Original post by ghostwalker)
From the context, I suspect it was just a typo, and they meant
That would make sense. Didn't notice that they might've meant this.
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1 year ago
#12
(Original post by Deranging)
This is a bit of a silly question really isn't it? Why would you do part (a) if you can just do part (b) and then repeat your answer :P.
Well, it's fairly easy to do (a) without much calculation:

Spoiler:
Show

Use the matrix that maps (x, y) to (0, x).

It's a little harder to find a more general matrix, although I agree that since you're going to have to do it, there's some truth to your argument that you should just do that and let (a) come out for free.

However, the idea that lets you write down the matrix for (a) should also let you do (b) without relatively little calculation.
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