# (CS 267, Mar 23 1995)

Theorem 3. (Special case of the Courant Fischer Minimax Theorem). If A is a symmetric matrix with eigenvalues lambda1 <= lambda2 <= ... lambdan and corresponding eigenvectors v1, v2, ..., vn then
```       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