lambdaand the minimizing v is v_{2}= min_{v != 0, v'*v_{1}= 0} v'*A*v / v'v

*Proof.* Substitute the eigendecomposition A = Q*Lambda*Q', where
Q is a orthogonal matrix whose columns v(i) are eigenvectors, and
Lambda = diag(lambda_{1}, ..., lambda_{n}) is a diagonal matrix of eigenvalues,
into the expression in the theorem:

v'*A*v minIt is easy to see that this expression is minimized by taking y = [0,1,0,...,0]', yielding lambda_{{v!=0, v'*v1 = 0}}------ v'v v'*Q*Lambda*Q'*v = min_{{v!=0, v'*v1 = 0}}---------------- v'*v v'*Q*Lambda*Q'*v = min_{{Qv!=0, v'*Q*Q'*v1 = 0}}---------------- v'*Q*Q'*v y'*Lambda*y = min_{{y!=0, y'*y1 = 0}}----------- y'*y where y = Q'*v and y_{1}= Q'*v_{1}= [1,0,...,0]' lambda_{i}*y(i)^{2}= min_{{y!=0, y(1) = 0}}sum_{i}------------- sum_{i}y(i)^{2}