reinout-g: Yet again I have stumbled upon a proof I seem to be unable to figure out.
Let matrix S be defined as a symmetric matrix.
Prove: tr(S) equals the sum of eigenvalues of S
Hope someone can help me out
Hi reinout-g!
This is true for any matrix.
Proof.
The eigenvalues are the solutions of det(S - tI) = 0.
This can be written both as:
(-1)^n ( t^n - tr(S) t^(n-1) + ... + (-1)^n det(S) ) = 0
and
(-1)^n (t - t_1)(t - t_2)...(t - t_n) = 0
Compare coefficients to see that:
tr(S) = t_1 + t_2 + ... + t_n
As a bonus, you can also see that:
det(S) = t_1 t_2 ... t_n
For more details of this proof see for instance page 3 in:
http://www.adelaide.edu.au/mathslearning/play/seminars/evalue-magic-tricks-handout.pdf
To be fair, this is not really a proof to come up with yourself.
I didn't.
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