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- April 2nd 2013, 01:54 PM #1

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- April 5th 2013, 03:37 AM #2

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## Re: Please help

Consider the matrices with a single 1 on the diagonal, 0 elsewhere. There are of these. There are also matrices with two 1s in symmetric positions off diagonal and with a 1 and -1 in symmetric positions off diagonal (so the resultant matrix is antisymmetric).

These form a basis for (they are clearly linearly independent, and there are of them). The symmetric matrices we have defined has eigenvalue 1 under , and the antisymmetric matrices we have defined has eigenvalue -1. It should be obvious what the matrix of looks like with respect to this basis: 1s and -1s down the diagonal, zero elsewhere.

The determinant of is the product of eigenvalues: