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Let A and B be matrices d x d with A Positive semi-definite and B positive-definite, and b a vector of Rd

1) Show that $$max_a \frac{(a^Tb)^2}{a^TBa} = b^TB^{-1}b $$

with the maximum is reached for all vector a=cB-1b, with c ϵ R.

Hint : use Cauchy-Schwartz inequality Inequality of Cauchy-Schwartz (extension) : Let a and b be two any vectors of Rd, and B a matrix d x d positive-definite. Then we have the following results :

inequality of Cauchy-Schwartz : $$ (a^Tb)^2 ≤ (a^Ta)(b^Tb) $$ an extension of this inequality gives : $$ (a^Tb)^2 ≤ (a^TBa)(b^TB^{-1}b)$$

I can also prove this:

$$max_a \frac{(a^Tb)^2}{a^TBa} = b^TB^{-1}b ⇔ max_a B^{-1}Aa = \lambda a$$under the constraint $$a^TBa = \lambda$$

I think I need to use the lagrangian.

2) Show that
$$max_a \frac{a^TAa}{a^TBa} = \lambda_1 $$

where λ1 is the greatest eigenvalue of B-1A. Conclude that the maximum is reached for the eigenvector B-1A which is associated to λ1

Rodrigo de Azevedo
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Tataria
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1 Answers1

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Hints:

  1. Cauchy-Schwarz tells you that $(x^Ty)^2\le (x^Tx)(y^Ty)$ and the equality attains when $x\|y$. Now apply it for $x=B^{1/2}a$, $y=B^{-1/2}b$. Here $B^{1/2}$ is the unique positive definite square root of $B$.
  2. For the same $x$ as above you get the Rayleigh quotient $$ \max_{a}\frac{a^TAa}{a^TBa}=\max_{x}\frac{x^TB^{-1/2}AB^{-1/2}x}{x^Tx}=\lambda_\max(B^{-1/2}AB^{-1/2}), $$ but matrices $B^{-1/2}AB^{-1/2}$ and $AB^{-1}$ are similar.
A.Γ.
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  • I don't really get it, how can I find the max with x=B1/2a, y=B−1/2b.... my teacher has given a hint that we had to use lagrange to maximize under the constraint that : B^-1 A a = λa – Tataria Oct 22 '17 at 03:02
  • @Tataria If you are talking about part 1 then $(a^Tb)^2\le (a^TBa)(b^TB^{-1}b)$ for all $a$ together with $(a^Tb)^2=(a^TBa)(b^TB^{-1}b)$ for some $a$ means trivially that $\max\frac{(a^Tb)^2}{a^TBa}=b^TB^{-1}b$ without any use of Lagrange functions. In part 2 one may use the Lagrange function, but I do not quite see why it is necessary. The optimization on $x$ is immediate after unitary diagonalization. Of course, it depends on what of matrix theory you may or may not use in your solution. – A.Γ. Oct 22 '17 at 11:10