Extremum of a multidimensional quadratic function

Question

I have the following function: $$ g(h) = h'\Sigma\Sigma'h-h'm-r, $$ where $h$ is a vector in $\mathbb{R}^M$, $\Sigma$ is a $M\times K$ matrix such that $\Sigma\Sigma'$ is positive definite and has full rank, $m$ is a vector and $r$ is a number. What is the minimum of this function? I tried to calculate FOC, but got lost in the notation...

Answer 1

We have, for $h, k \in \mathbb R^M$: \begin{align*} g(h+k) &= (h+k)^t \Sigma\Sigma^t(h+k) - (h+k)^tm - r\\ &= h^t\Sigma\Sigma^t h + 2h^t\Sigma\Sigma^t k + k^t \Sigma\Sigma^t k - h^tm - m^tk -r \\ &= g(h) + (2h^t\Sigma\Sigma^t - m^t)k + k^t \Sigma \Sigma^t k \end{align*} As, $k^t \Sigma \Sigma^t k = o(\lvert k \rvert)$, $g$ is differentiable with $g'(h)k = 2h^t\Sigma\Sigma^tk-m^tk$, any $h,k \in \mathbb R^M$, we have $g'(h) = 0$ exactly for $h = (2\Sigma\Sigma^t)^{-1}m$ and as $g''(h) = 2\Sigma\Sigma^t$ is positive, it is a minimum.