Digression: An application of Lagrange multipliers to spectral theory

The following fact about eigenvalues has been useful in previous lectures: If 144#18 is a symmetric matrix, then

145#19

Dividing through by 146#20, this result can be written as

 147#21 (12)

I will now show that, in fact,

148#22

and

149#23

I define $f:{\bf {\rm R}}^n\rightarrow{\bf {\rm R}}$ by

150#24

and 151#25 by

152#26

I can then find the maximum and minimum of f subject to the constraint g(x)=0. The gradients are

153#27

and therefore the Lagrange multiplier condition is

154#28

The stationary points are the eigenvectors 155#29of A, and, with 156#30,

157#31

Therefore the maximum value of 158#32, subject to 159#33, is 160#34, and the minimum value is 161#35.