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Let, $\mathbf{A}$ be a symmetric positive definite matrix. I want to minimize $$f(x) = \frac{1}{2}x^T \mathbf{A}x - b^{T}x.$$ How would I go about this? I think I have to recast this as a least squares problem.

Johndoe
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1 Answers1

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Differentiate the expression as it was done here. You end up with the jacobian of $f$:

$$x^\top \mathbf{A} -b^\top. $$ To find the minimum, the expression above must be equal to $0$, and so we find $$x^\top \mathbf{A} = b^\top.$$ $\mathbf{A}$ is positive definite and thus invertible, and so we find an extremum of the function at $x = \mathbf{A}^{-1} b$. Finally, differentiate the first expression again to obtain the Hessian of $f$, i.e. $\mathbf{A}$. Since $\mathbf{A}$ is positive definite, the extremum you obtained is a minimum of your function.

Pengin
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