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Consider the vector z. The problem is:

$\max {z^T x}$, subject to $\ x^T P x \leq 1$, where $P$ is a positive definite symmetric matrix.

Show that the optimal value is $\sqrt{z^TP^{-1}z}$. Once this is proof use this to establish the inequaity:

$\ (x^Tz)^2 \leq (x^TPx)(z^TP^{-1}z)$.

I am having problems to proof the inequality.

Ptr34543
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2 Answers2

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Define $|||x|||=\sqrt{x^TPx}$. The optimization constraint is then $|||x|||\le 1$. The vector $$ \hat x=\frac{x}{|||x|||} $$ satisfies the constraint ($|||\hat x|||=1$), hence, $$ z^T\hat x\le\max z^Tx=\sqrt{z^TP^{-1}z}. $$ We are done after some rearranging the inequality.

A.Γ.
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Hint: Try rewriting the equation you derived as follows:

$$ (z^T x)(z^T x) = z^T P^{-1} z $$

Then, transpose each side of this expression.

You should then be able to manipulate the expression to look like the inequality you are trying to obtain. From there, you will need to think about the properties of the matrices provided in the statement of the problem and use these to complete the proof.

Chris
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  • If I make the transpose I get $\ (x^Tz)(x^Tz)=(x^Tz)^2=z^T P^{-1} z$ as a maximum, but I wonder how to introduce the term $\ x^T P x$ – Ptr34543 Nov 11 '18 at 14:24
  • Think about how $z^T P^{-1} z$ compares to $(x^T P x)(z^T P^{-1} z)$. Specifically, think about the properties of $P$ and what these imply. – Chris Nov 12 '18 at 03:32