Concave function with Hessian I need to prove that this function $S(E,V,N)=k\cdot \:N\cdot \:ln\left(\frac{V}{N\cdot \:v_0}\right)+\frac{3}{2}\cdot \:k\cdot \:N\cdot \:ln\left(\frac{E}{N\cdot \:e_0}\right)$ is concave (with N number of molecules, E the energy, and V the volume). I have calculated the Hessian:
$\begin{pmatrix}\frac{3k}{2E}&\frac{k}{V}&-\frac{5k}{2N}\\ \:-\frac{3kN}{2E^2}&0&\frac{3k}{2E}\\ \:0&-\frac{kN}{V^2}&\frac{k}{V}\end{pmatrix}$ and the determinant is = $0$, so now I want to prove that the Hessian is everywhere negative semi-definit, but I cannot calculate the eigenvalues, so I have thought that on a certain 2d subspace it is negative definite, but I can´t find that subspace.
 A: Firstly, in any case Hessian shall be a symmetric matrix, so I think your Hessian matrix has its row permuted...
Using mathematica to compute it, the Hessian appears to be this (take $k=1$)
$$
H=\left(
\begin{array}{ccc}
 -\frac{3 N}{2 \text{E}^2} & 0 & \frac{3}{2 \text{E}} \\
 0 & -\frac{N}{V^2} & \frac{1}{V} \\
 \frac{3}{2 \text{E}} & \frac{1}{V} & -\frac{5}{2 N} \\
\end{array}
\right)
$$
Real symmetric matrices are diagonlizable by orthonormal matrices, and it has real eigenvalues.
$$
\lambda_{1,2,3}=0,-\frac{\alpha\pm\sqrt{\beta}}{4 \text{E}^2 N V^2}\\
\alpha =\text{E}^2 \left(2 N^2+5 V^2\right)+3 N^2 V^2\\
\beta=4 \text{E}^4 N^4-4 \text{E}^2 N^2 V^2 \left(\text{E}^2+3 N^2\right)+V^4 \left(25 \text{E}^4+6 \text{E}^2 N^2+9 N^4\right)
$$
By brute force calculation you got
$$
\alpha^2-\beta=24 \text{E}^2 N^2 V^2 \left(\text{E}^2+N^2+V^2\right) >0\\
\alpha>\sqrt\beta
$$
Thus
$$
\lambda_2<0,\lambda_3<0
$$
The Hessian matrix is negative semidefinite. The null space is $[\frac{\text{E}}{N},\frac{V}{N},1]^T$, which makes sense physically: This direction corresponds to scaling up $E,N,V$ the whole system by the same coefficient. Since $S$ is a linear function of scale, it has 0 curvature along this dimension, i.e. the Hessian has 0 eigenvalue along this direction.
