# Using bordered Hessian matrix to determine non-degeneracy and type of constrained extremum

I have the following problem: $\def\f{f(x_1,x_2,x_3)}\def\1{x_1}\def\2{x_2}\def\3{x_3}\def\n{\nabla}\def\g{g(x_1,x_2,x_3)}\def\l{\lambda}\def\q{\begin{pmatrix}}\def\p{\end{pmatrix}}$

Find the critical points of the following constrained optimization problem: $$\f=\1^2+\2^2+2\3^2$$ subject to $$\g=\1+\2+\3=4$$

and am required to check that these are non-degenerate and need to determine local min and maxima.

$$L = \f+\l \g$$ $$=\1^2+\2^2+2\3^2+\l\1+\l\2+\l\3-4$$ $$\n L= (2\1+\l,2\2+\l,4\3+\l)=\vec{0}$$ Which yields: $$\1+\2+\3=4 \;\; (1)$$ $$2\1+\l=0 \;\;(2)$$ $$2\2+\l=0\;\;(3)$$ $$4\3+\l=0\;\;(4)$$

From 2),3), $\1=\2$,from 4),$\3=\frac{\2}2$

From 1) $\frac52\2=4\Rightarrow \2=\frac85$

$$\1=\frac85=\2,\3=\frac8{10},\l=-5$$

$f(\frac85,\frac85,\frac8{10})=\frac{32}{5}$

To check degeneracy, I have to look at the bordered Hessian matrix.

Can someone show me how to use this bordered Hessian matrix?

My attempt: $$H = \q 0 & g_{\1} & g_{\2} & g_{\3} \\ g_{\1}&L_{{\1}{\1}}&L_{{\1}{\2}}&L_{{\1}{\3}} \\ g_{\2}&L_{{\2}{\1}}&L_{{\2}{\2}}&L_{{\2}{\3}} \\ g_{\3}&L_{{\3}{\1}}&L_{{\3}{\2}}&L_{{\3}{\3}} \p$$

$$H = \q 0 & 1 & 1 & 1 \\ 1&2&0&0 \\ 1&0&2&0 \\ 1&0&0&4 \p$$

Taking the determinant of this yields $H = -14 \ne 0$ Hence this is not a degenerate point.

• See this answer which has correct bordered Hessian for similar functions. – user147263 Aug 24 '14 at 3:04
• @Thursday Thank you for that, if possible, could you verify my answer? – Tony Aug 24 '14 at 3:25
• I don't think that saying $H\ne 0$ is enough; the bordered Hessian requires consideration of upper-left minors of certain sizes, as described in Bordered Hessian article. Here $m=1$ and $n=3$, so you check the signs of $3\times 3$ minor (negative) and the full matrix (also negative). Conclusion: minimum. ... For the future: observe that searching can lead to useful information faster than typing the question. – user147263 Aug 24 '14 at 3:42
• @Thursday searching can lead to useful information faster, but verification it (usually) cannot. – Tony Aug 24 '14 at 3:45
• @Thursday Thank you for title edit. I will refer to your helpful link and will in future refer to the search function more readily. – Tony Aug 24 '14 at 3:48

Expand at the first line $\det H = -\det\begin{pmatrix}1 &0 & 0\\ 1 & 2 & 0 \\ 1 &0 & 4\end{pmatrix}+\det\begin{pmatrix}1 &2 & 0\\ 1 & 0 & 0 \\ 1 &0 & 4\end{pmatrix}-\det\begin{pmatrix}1 &2 & 0\\ 1 & 0 & 2 \\ 1 &0 & 0\end{pmatrix}=-8-8-4=-20$
• I see, I accidently took $-1*4 + 4*[-2-2]=-20$ as $-1*4 + 4*-2 -2=-14$ – Tony Aug 24 '14 at 3:43