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I'm solving constrained optimization exercises for preparing my final exam. I got stuck at this question.

$$ \begin{array}{ll} \text{min} & \mathbf{x}^T\mathbf{A}\mathbf{x} \\ \text{s.t.} & \|\mathbf{x}\|_2=1 \end{array} $$

I'm asked to show that, if the minimum point of the function is $\mathbf{x}^*$ in the domain, then magnitude of the smallest eigenvalue of $\mathbf{A}$ is ${\mathbf{x}^*}^T\mathbf{A}\mathbf{x}^*$.

I tried these:

$$ \begin{array}{lcll} f(\mathbf{x}) & = & \mathbf{x}^T\mathbf{A}\mathbf{x} & \qquad \text{(objective function)} \\ h(\mathbf{x}) & = & \mathbf{x}^T\mathbf{x}-1 = 0 & \qquad \text{(equality constrain)} \end{array} $$

The Lagrangian is

$$ \mathcal{L}(\mathbf{x},v) = f(\mathbf{x}) + vh(\mathbf{x}) = \mathbf{x}^T\mathbf{A}\mathbf{x} + v\mathbf{x}^T\mathbf{x} - v, $$

where $v$ is the Lagrange multiplier.

Next, I tried finding infinimum of $\mathcal{L}(\mathbf{x},v)$.

$$ \mathbf{0} = \dfrac{\partial}{\partial \mathbf{x}} \mathcal{L}(\mathbf{x},v) = \mathbf{x}^T(\mathbf{A}+\mathbf{A}^T) + 2v\mathbf{x}^T \\ (\mathbf{A} + \mathbf{A}^T)\mathbf{x}^* + 2v^*\mathbf{x}^* = \left[(\mathbf{A} + \mathbf{A}^T) + 2v^*\mathbf{I}\right]\mathbf{x}^* = \mathbf{0} \\ \implies v^*\mathbf{I} = -\dfrac{1}{2}(\mathbf{A} + \mathbf{A}^T) \qquad \text{(doesn't seem to be correct)} $$

I'm not able to go on any further. How do I find $\mathbf{x}^*$ after this point?

Rodrigo de Azevedo
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hkBattousai
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2 Answers2

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I will write the Lagrangian as objective -$v$ constrain (rather than +, but they are both correct) $$ (\mathbf{A} + \mathbf{A}^T)\mathbf{x}^* - 2v\mathbf{x}^* = \left[(\mathbf{A} + \mathbf{A}^T) - 2v\mathbf{I}\right]\mathbf{x}^* = \mathbf{0} $$ does not imply $$ v\mathbf{I} = \dfrac{1}{2}(\mathbf{A} + \mathbf{A}^T) \qquad $$

Rather it implies $$det((\mathbf{A} + \mathbf{A}^T) - 2v\mathbf{I})=0,$$ because it is a vector equation not scalar.

And this is the definition of characteristic function for eigenvalues. So that tells us the dual variable $v$ is an eigenvalue of $A$ when duality gap is zero or when the primal and dual pair exists.

Now to show that $\mathbf{x}^*\mathbf{A}^T\mathbf{x}^* $ is the smallest eigenvalue we need to assume that $A$ is positive definite. I think this must be given as otherwise the optimization problem is not convex and hence we won't be able to find a unique $\mathbf{x}^*$. Assuming unique solution and from $\mathbf x^*$ and $v$ being the eigenvector and eigenvalue note that we have $$A\mathbf{x}^*=v\mathbf x^*$$ then$$\mathbf{x}^{*T} A\mathbf{x}^*=v\mathbf x^* \mathbf x^{*T}$$ and since $\mathbf x^* \mathbf x^{*T}=1$ $$\mathbf{x}^{*T} A\mathbf{x}^*=v$$ and since we minimized $$\mathbf{x}^T A\mathbf{x}$$ we have that $\mathbf{x}^{*T} A\mathbf{x}^*$ indeed is the minimum eigenvalue.

triomphe
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  • Could you explain why it is that the determinant must be $0$? Could it not happen than $x^*$ belongs to the null space of $(A+A^T)-2vI$ instead? – peteykun Dec 11 '16 at 07:30
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The first equation where you set the derivative to zero, is in fact describing an eigenvalue and the corresponding eigenvector of $\frac{1}{2}(A+A^T)$. So the stationary points of the Lagrangian are eigenvectors.

One correction: In your second to last equation you should have $\cdots+2v^*I$ rather than $v^*$ (note that $v^*$ is a scalar not a matrix). However, you don't need this equation at all. The first one suffices.

Also I think the solution to the minimization is the smallest eigenvalue of $\frac{1}{2}(A+A^T)$ rather than that of $A$. If $A$ is symmetric then both of these values coincide.

S.B.
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