lambda(2) = min_{v != 0, v'*v(1) = 0} v'*A*v / v'vand the minimizing v is v(2).
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:
min_{v != 0, v'*v(1) = 0} v'*A*v / v'v = min_{v != 0, v'*v(1) = 0} v'*Q*Lambda*Q'*v / v'*v = min_{Qv != 0, v'*Q*Q'*v(1) = 0} v'*Q*Lambda*Q'*v / v'*Q*Q'*v = min_{y != 0, y'*y(1) = 0} y'*Lambda*y / y'*y where y = Q'*v and y(1) = Q'*v(1) = [1,0,...,0]' = min_{y != 0, y_1 = 0} sum_i lambda(i)*y_i^2 / sum_i y_i^2It is easy to see that this expression is minimized by taking y = [0,1,0,...,0]', yielding lambda(2). Then v = Q*y = v(2), as desired. QED