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min $\left \langle a,x \right \rangle$ subject to $\left \| x \right \|^2 \le 1$

What was done:

$L(x,\lambda) = a^Tx + \lambda^T(x^Tx-1) = a^Tx + \lambda^Tx^Tx-\lambda^T$

that is, $L(x,\lambda) = inf\{(a^T+\lambda^Tx^T)x\}-\lambda^T$

Gradients are:

$\nabla f(x) = a$ e $\nabla g(x) = 2x$

Therefore, $\nabla L(x,\lambda) = \nabla f(x) + \lambda \nabla g(x) = 0$

$x^* = \frac{-a}{2\lambda}$

Replacing $x^*$ em $L(x, \lambda)$ we have:

$(a^T + \lambda^T(\frac{-a}{2\lambda})^T)(\frac{-a}{2\lambda}) - \lambda^T$

My doubt is, that when I do this distributivity I end up not seeing the dual.

Amissadai ferreira
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1 Answers1

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  1. $<a,x>$ is $a^Tx$ not $ax^T$

  2. $\lambda$ is a scalar so no point in writing $\lambda^T$

  3. When replacing $x^*$ you forget multiplying it with $a^T$

  4. You are incorrectly replacing $x^Tx$

Vishaal Sudarsan
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  • I edited my solution according to your answer, I just left the lambda transposed, because it is a vector of n dimension. Could you verify that it is now correct? Sorry for the writing, because it is not my mother tongue. – Amissadai ferreira Nov 14 '20 at 00:09
  • Continuing $(\frac{-a}{2\lambda})a^T + (\frac{-a}{2\lambda})^T\lambda^T(\frac{-a}{2\lambda}) - \lambda^T$ – Amissadai ferreira Nov 14 '20 at 00:11
  • $\left \langle a,-\frac{a}{2\lambda} \right \rangle + \left \langle \lambda^T(-\frac{a}{2\lambda}),-\frac{a}{2\lambda} \right \rangle - \lambda^T$ – Amissadai ferreira Nov 14 '20 at 00:16
  • $-\left \langle a,\frac{a}{2\lambda} \right \rangle - \left \langle \lambda^T(\frac{a}{2\lambda}),\frac{a}{2\lambda} \right \rangle - \lambda^T$ – Amissadai ferreira Nov 14 '20 at 00:19
  • $\lambda$ is $\textbf{not}$ a vector. There is only one constraint in this optimization problem hence it’s just a scalar. – Vishaal Sudarsan Nov 14 '20 at 10:48