Why is $\max_{c^TBc=1}c^T a a^Tc =a^TB^{-1}a$?

Question

I'm having some trouble in understanding the last step in this sequence of equalities:

$$\max_{c}\frac{c^Taa^Tc}{c^TBc}=\max_{c^TBc=1}c^T a a^Tc =a^TB^{-1}a$$

I would think that the maximum would be equal the biggest eigenvalue of the matrix $B^{-1}aa^T$

Answer 1

I assume $B$ is positive definite. Defining $x=B^{1/2}c$ the problem is $$ \max_{c^TBc=1}|a^Tc|^2=\max_{\|x\|=1}|a^TB^{-1/2}x|^2=\|B^{-1/2}a\|^2=a^TB^{-1}a. $$

EDIT: the eigenvalues of $B^{-1}aa^T$ are the same as the eigenvalues of the (similar) matrix $B^{-1/2}aa^TB^{-1/2}$, which has the same nonzero eigenvalues as the matrix (actually, number) $a^TB^{-1/2}B^{-1/2}a=a^TB^{-1}a$. Your conclusion is right: the maximum is the largest eigenvalue of $B^{-1}aa^T$, and it is $a^TB^{-1}a$.

Answer 2

You can also use Lagrange multiplier to show it, i.e., you problem is equivalent to $$ \arg \max_{c, \lambda} \left( c'Ac - \lambda (c'Bc - 1) \right), $$ where $A = aa'$. The first order condition is $$ 2Ac - 2\lambda Bc=0, $$ i.e., $$ Ac=\lambda Bc, $$ that is a generalized eigenvalue problem. If $B$ is invertible, then $$ B^{-1}Ac=\lambda c, $$ that is the vector $c$ that satisfies the F.O.C, must satisfy this linear, which is by the definition pairs of eigenvectors and eigenvalues of $B^{-1}A$. As such, the final question is which pair to choose? So, going back to the original formulation you have in the nominator $$ c'Ac=c' \lambda_A c = \lambda_A \sum_{i=1}^nc_i^2, $$ where the pair of largest eigenvalue and eigenvector of $A$ correspond to the same pair in $B^{-1}A$.

Alternatively you can use even more basic calculus to show it. Recall that $$ \left( \frac{f(x)}{g(x)} \right)' = \frac{f'g - fg'}{ g^ 2 }, $$ hence the F.O.C is, $$ \left(\frac{c'Ac}{c'Bc} \right)' = \frac{2Ac c'Bc - 2c'AcBc}{(c'Bc)^2} = 0, $$ rewriting the LHS and multiplying both sides by $c'Bc/2$ you have $$ Ac - Bc\frac{ c'Ac }{c'Bc} = 0, $$ Note that $\frac{ c'Ac }{c'Bc}$ is a (positive) scalar, so denote it by $\lambda$, thus you'll get the generalized eigenvalue problem, i.e., $$ Ac = \lambda B c, $$ hence as previously you have $$ B^{-1}Ac = \lambda c, $$ i.e., the mximum is attained for $c$ that is eigenvector of $B^{-1} A$ and the maximum value equals its largest eigenvalue.