lambda2 = min_{v != 0, v'*v1 = 0} v'*A*v / v'v
and the minimizing v is v2.
Proof. Substitute the eigendecomposition A = Q*Lambda*Q', where Q is a orthogonal matrix whose columns v(i) are eigenvectors, and Lambda = diag(lambda1, ..., lambdan) is a diagonal matrix of eigenvalues, into the expression in the theorem:
v'*A*v
min{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 y1 = Q'*v1 = [1,0,...,0]'
lambdai*y(i)2
= min{y!=0, y(1) = 0} sumi -------------
sumi y(i)2
It is easy to see that this expression is minimized by taking
y = [0,1,0,...,0]', yielding lambda2. Then v = Q*y = v2, as desired. QED