Is there an upper bound on the trace of positive definite matrix as a function of the trace of the inverse of the matrix?

Question

Let $A$ be an $n \times n$ positive definite matrix. In this answer (based on a question I previously asked), a lower bound is given on $\text{tr}[A]$ as a function of $\text{tr}[A^{-1}]$:

$$\text{tr}[A] \geq \dfrac{n^2}{\text{tr}[A^{-1}]}$$

Is there also an upper bound? I can think of an upper bound on the trace as a function of the maximum eigenvalue of $A$. For example,

$$ \text{tr}[A] \leq n \lambda_{\max}[A] $$

However, I can't think of an upper bound as a function of the trace of the inverse of $A$.

Answer 1

The relationship between the trace of $A$ and the trace of $A^{-1}$ must depend on additional parameters other than $n$. Consider the example: $$ A = \begin{bmatrix} 1 & 0 \\ 0 & \delta \end{bmatrix}, $$ where $\delta > 0$ is small. We observe that $\mathrm{tr}(A) = 1 + \delta$, $\mathrm{tr}(A^{-1}) = 1 + 1/\delta$, so $\mathrm{tr}(A)$ is nearly constant for small $\delta$ while $\mathrm{tr}(A^{-1})$ blows up.

The missing ingredient needed is the condition number. Generally speaking, the relationship between $A$ and $A^{-1}$ can be often characterized by the value of the condition number $\kappa$. For a general $n \times n$ matrix $A$, the condition number is defined by $$ \kappa(A) = \frac{\sigma_{max}(A)}{\sigma_{min}(A)}, $$ where $\sigma_{max}$ and $\sigma_{min}$ are the largest and smallest singular values. If $A$ is invertible then this is equal to $\|A\| \|A^{-1}\|$. For an SPD matrix, this is equivalent to $\lambda_{max}(A)/\lambda_{min}(A)$.

Suppose $\kappa$ is large. Then, rescaling $A$ to have $\sigma_{max} = 1$, this means that $A$ has small norm but $A^{-1}$ has a large norm. This means that $A$ has a small trace, yet $A^{-1}$ has a large trace. Conversely, if $\kappa$ is close to $1$, then $A$ and $A^{-1}$ has similar norms, and their traces are also similar.

Going back to our original example, $\kappa(A) = 1/\delta$. Because the condition number of this matrix is large, then the trace of $A$ is close to $1$ while the trace of $A^{-1}$ is very large.

Now we can address your question. We have \begin{align} \mathrm{tr}(A) & = \lambda_1 + \lambda_2 + \ldots + \lambda_n \\ \mathrm{tr}(A^{-1}) & = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \ldots + \frac{1}{\lambda_n}. \end{align}

Assuming $\lambda_1 \geq \ldots \geq \lambda_n > 0$, and utilizing the inequalities $$ \frac{\lambda_1}{\kappa} \leq \lambda_j \le \lambda_n \kappa, $$ we have \begin{align} \mathrm{tr}(A) \mathrm{tr}(A^{-1}) & \ge \left( \frac{ n \lambda_1}{\kappa} \right) \left( \frac{n}{\kappa \lambda_n} \right) = \frac{n^2}{\kappa}, \\ \mathrm{tr}(A) \mathrm{tr}(A^{-1}) & \le \left(n \lambda_n \kappa \right) \left( \frac{n \kappa}{ \lambda_1} \right) = n^2 \kappa. \end{align} In summary we have the inequalities $$ \boxed{\frac{n^2}{\kappa} \le \mathrm{tr}(A) \mathrm{tr}(A^{-1}) \le n^2 \kappa} $$ which can be converted into lower and upper bounds on the trace of $A$, in terms of the trace of $A^{-1}$, the condition number $\kappa$, and $n$.

Answer 2

Unless $n=1$, there is no such upper bound. For example, let $\epsilon>0$ and $$ A=\pmatrix{1+\frac{1}{\epsilon}\\ &1+\epsilon}. $$ Then $\operatorname{tr}(A^{-1})=1$ but $\operatorname{tr}(A)\to\infty$ as $\epsilon$ approaches $0$ or infinity.