1) The proof for #10 under eigenvalues (discussed during the first recitation) requires the assumption that A is symmetric in addition to idempotent. Actually as long as the matrix of eigenvectors is invertible the proof works, but I haven't seen a working proof for the result with solely A is idempotent assumption. 2) A student latexed a proof of Property 5 (Rank(A) = # of nonzero eignvalues) which was better than my psuedoproof. Thanks! The proof can be found at the website. https://sites.google.com/site/enochhill/home/teaching/econ-8205-fall-2011 Also it was pointed out that this should more properly be stated as the Rank(A) = # of nonzero eigenvalues counting multiplicity. 3) One student asked if to prove "sum of eigenvalues is equal to the trace" if we need the assumption that the matrix is diagonalizable, Another student responded: I think we don't need to assume the matrix is diagonalizable. If we look at the characteristic polynomial of the matrix: p(lambda)=det(lambda*I-A)=(lambda-lambda_1)*(lambda-lambda_2)*...*(lambda-lambda_n) sum of eigenvalues is equal to (-1)^n*(coefficient of lambda^(n-1)), which is exactly equal to the trace of A.
1) The proof for #10 under eigenvalues (discussed during the first recitation) requires the assumption that A is symmetric in addition to idempotent. Actually as long as the matrix of eigenvectors is invertible the proof works, but I haven't seen a working proof for the result with solely A is idempotent assumption.
ReplyDelete2) A student latexed a proof of Property 5 (Rank(A) = # of nonzero eignvalues) which was better than my psuedoproof. Thanks! The proof can be found at the website.
https://sites.google.com/site/enochhill/home/teaching/econ-8205-fall-2011
Also it was pointed out that this should more properly be stated as the Rank(A) = # of nonzero eigenvalues counting multiplicity.
3) One student asked if to prove "sum of eigenvalues is equal to the trace" if we need the assumption that the matrix is diagonalizable,
Another student responded: I think we don't need to assume the matrix is diagonalizable. If we look at the characteristic polynomial of the matrix:
p(lambda)=det(lambda*I-A)=(lambda-lambda_1)*(lambda-lambda_2)*...*(lambda-lambda_n)
sum of eigenvalues is equal to (-1)^n*(coefficient of lambda^(n-1)), which is exactly equal to the trace of A.
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