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#1
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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? |
#2
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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 |
#3
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Re: Linear algebra, matrices, and you
Dude, it's a bonus problem.
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#4
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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. |
#5
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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 |
#6
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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. |
#7
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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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#8
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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. |
#9
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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. |
#10
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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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