# Is maximizing the trace of the inverse of a positive definite matrix with eigenvalues greater than $1$ the same as minimizing the determinant?

Lets say I want to minimize the inverse of the following matrix: $$A = \begin{bmatrix} \lambda_1 &0 & 0\\ 0& \lambda_2 &0\\0& 0& \lambda_3 \end{bmatrix}$$ where all $\lambda \geq 1$. I also read that the determinant of the a the matrix $A$ can be given by: $$\det(A) = \lambda_1 \lambda_2 \lambda_3$$. The trace of inverse of $A$ is given by: $$trace(A^{-1}) = \frac{1}{\lambda_1}+\frac{1}{\lambda_2}+\frac{1}{\lambda_3}=\frac{\lambda_1 \lambda_2 + \lambda_2 \lambda_3 + \lambda_1 \lambda_3}{\lambda_1 \lambda_2 \lambda_3}$$

So for $\lambda \geq 1$, is it safe to say that minimizing the trace of inverse of of A is the same as maximizing the determinant of A?

• What about the other entries ? Is A a diagonal matrix ? Or are just the entries below the diagonal zeros ? Please be more precise! – Peter May 8 '17 at 17:19
• @RobertIsrael Trace of inverse – Anon123 May 8 '17 at 17:22
• Minimizing or maximizing subject to what constraints? If your only constraint is $\lambda \ge 1$, then the determinant can be arbitrarily large, and the trace of $A^{-1}$ can be arbitrarily close to $0$. – Robert Israel May 8 '17 at 17:23
• @Peter It is a diagonal matrix. In general trace is always the sum of eigenvalues, no? – Anon123 May 8 '17 at 17:23
• I think investigating matrix logarithm may be more fruitful. Then also starting from $I$ ($\lambda_k = 1 , \forall k$) becomes more natural (why?). – mathreadler May 8 '17 at 17:27

## 1 Answer

This is not equivalent. Let $A_1 = diag(1,2,2)$ and $A_2 = diag(1,1,5)$. Then $$trace(A_1^{-1}) = 2 < trace(A_2^{-1}), \quad det(A_1) = 4 < det(A_2) \, .$$