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  #1  
Old 04-21-2006, 03:21 PM
TBag TBag is offline
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Default Linear algebra, matrices, and you

So yes, here is a math problem I'm rather curious about. In this situation, I'm guessing that A is just an identity matrix of n x n, would that be correct? If so, how would one go about proving it?



Also, for the true or false questions I got
1 true
2 false
3 I'm pretty sure this one is true, because if C = A^-1, and B = C^-1, therefore A=B, but I'm not sure I'll work one out and then edit my answer in
4 True

edit - I did number 3 with the matrix
1 2
3 4
And it proved the statement true. Is there any instance where it wouldn't?
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  #2  
Old 04-21-2006, 03:49 PM
brandofo brandofo is offline
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Default Re: Linear algebra, matrices, and you

[ QUOTE ]
So yes, here is a math problem that I need to finish by Monday. Please do it for me.

[/ QUOTE ]
FYP
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  #3  
Old 04-21-2006, 03:55 PM
TBag TBag is offline
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Default Re: Linear algebra, matrices, and you

Dude, it's a bonus problem.
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  #4  
Old 04-21-2006, 04:08 PM
thedustbustr thedustbustr is offline
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Default Re: Linear algebra, matrices, and you

[ QUOTE ]
I'm guessing that A is just an identity matrix of n x n

[/ QUOTE ]
No need to guess, this is given. rereading the question would probably be a good wy to get started.
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  #5  
Old 04-21-2006, 04:25 PM
TBag TBag is offline
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Default Re: Linear algebra, matrices, and you

But that's not necessarily true. If you have a 2 x 2 matrix, the possible matrices I see working in addition to the identity matrix are

Err, have to edit these made a mistake

-1 0
0 -1

0 1
1 0

0 -1
-1 0

For example, if your matrix is [0 -1 : -1 0], multiplying by it's transpose is

0(0) + (-1)(-1) (0)(-1) + (-1)(0)
0(-1) + (0)(-1) (-1)(-1) + (0)(0)

Which is [1 0 : 0 1], an identity 2 x 2 matrix. I dunno if I"m doing something wrong, but I don't think we can prove it's definately an identity matrix
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  #6  
Old 04-21-2006, 07:05 PM
jason1990 jason1990 is offline
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Default Re: Linear algebra, matrices, and you

If you regard all n-vectors as n x 1 matrices, then the dot product of two vectors, x and y, is just the matrix product, x^T y. This fact might be helpful.
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  #7  
Old 04-21-2006, 08:14 PM
TBag TBag is offline
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Default Re: Linear algebra, matrices, and you

So what about this,

since we're doing dot products and the matrix a is always a diagonal matrix of 1's or -1's, either going from top left to bottom right or bottom left to top right.
IE
[(-)1 0 0 : 0 (-)1 0 : 0 0 (-)1 ],
[0 0 (-)1 : 0 (-1) 0 : (-1) 0 0 ], etc

We can assume one of two things is happening.

1)
Ax = (A11 * x1) + (A22 * x2) ... (Ann * xn)
and same goes for y, replacing the x

or if the ones line up from bottom left to top right
2)
Ax = (A1n * xn) + (A2(n-1) * x(n-1) ) .... (An1 * x1)
and same goes for y,

so regardless of which path we take, in Ax (dot) Ay, we end up with (either A entry)^2(x1)(y1)+ ... + A^2(xn)(yn) which is the same as regular x (dot) y because any A entry squared will be 1.
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  #8  
Old 04-21-2006, 09:09 PM
cliff cliff is offline
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Default Re: Linear algebra, matrices, and you

A is an orthogonal matrix, which mean A A^T=I=A^T A, for instance
A=1/sqrt(2)* [1 1
1 -1]
in fact there are an infinite number of such matrices (in the complex field at least) and the set of all such matrices are closed under multiplication. They show up a lot in the theory of matrices.
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  #9  
Old 04-21-2006, 09:14 PM
cliff cliff is offline
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Default Re: Linear algebra, matrices, and you

A does not have to be diagonal (see my example of the Hadamard matrix in the above post), what you want here is that
Ax . By= B^T Ax . y for any matrices A and B by the definition of the inner product.
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  #10  
Old 04-21-2006, 11:33 PM
jason1990 jason1990 is offline
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Default Re: Linear algebra, matrices, and you

As cliff mentioned, A does not have to be diagonal. Frankly, I think you should stop trying to dig into the entries of the matrix A. Think more abstractly. Are you aware of the formula (AB)^T = (B^T)(A^T)?
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