Trace of an inverse matrix

Question

I want to know if there is a way to simplify or a closed form solution of $\operatorname{tr} \left( \Sigma^{-1} \right)$, where $\Sigma$ is a symmetric positive definite matrix.

Answer 1

Let $A$ be symmetric positive definite matrix hence $\exists$ a diagonal matrix $D$ whose diagonal entries are nonzero and $A=P D P^{-1}$ so $A^{-1} = P D^{-1} P^{-1}$ and $Tr(A^{-1})= Tr(D^{-1})$. Now $D$ being diagonal matrix with non zero diagonal entries $D^{-1}$ has diagonal entries reciprocal of the diagonal entries of $D$ so $Tr(D^{-1})$ is sum of the inverses of the diagonal entries of $D$.

Answer 2

It seems worth making explicit what I think Tobias means in his comment on the accepted answer: the characteristic polynomial of a matrix $A$ can be used to compute $\text{tr}(A^{-1})$ explicitly as follows:

Recall that the characteristic polynomial of an $n \times n$ matrix $A$ is defined to be the polynomial in one variable given by $$ \tag{$\dagger$} \chi_A(t) = \det(tI_n - A) = t^n + \sum_{i=0}^{n-1}(-1)^{n-i} c_i(A)t^i \in \mathbb R[t]. $$ where the coefficients $(-1)^{n-i}c_i(A)$ are thus polynomial functions of the entries of $A$. Indeed the constant term is just $$ (-1)^nc_0(A) =\det(-A)=(-1)^n \det(A) $$ Since $A$ is symmetric, $\mathbb R^n$ has an orthonormal basis $\{u_1,\ldots,u_n\}$ consisting of eigenvectors of $A$. It follows that there is an orthogonal matrix $P$ such that $P^\intercal AP = D$ where $D$ is the diagonal matrix $\text{diag}(\lambda_1,\ldots,\lambda_n)$, where $Au_i = \lambda_i u_i$, that is, $\lambda_i$ is the eigenvalue associated to the eigenvector $u_i$. But then using the multiplicativity of the determinant, we see that $$ \begin{split} \det(tI_n - D) &= \det(tI_n - P^\intercal AP) \\ &= \det(P^{\intercal}(tI_n - A)P) \\ &= \det(P^\intercal)\det(tI_n-A) \det(P)\\ &= \det(tI_n -A), \end{split} $$ and hence $$ \chi_A(t) = \det(tI_n- D) = \prod_{i=1}^n (t-\lambda_i) = t^n + \sum_{i=0}^{n-1} (-1)^{n-i}e_{n-i}(\lambda_1,\ldots, \lambda_n)t^{i} $$ so that the (signed) coefficients $c_i(A)$ are given by $c_i(A) = e_{n-i}(\lambda_1,\ldots,\lambda_n)$, the $i$-th elementary symmetric function evaluated at the eigenvalues $\lambda_1,\ldots,\lambda_n$.

Now as $$ \text{tr}(A) = \text{tr}(P(DP^{\intercal})) = \text{tr}((DP^{\intercal})P) = \text{tr}(D) = \sum_{i=1}^n \lambda_i, $$ we see that $c_{n-1}(A) = \text{tr}(A)$.

Now $$ \chi_A(t^{-1}) = \det(t^{-1}I_n-A) = \det((-t^{-1}A)(tI_n-A^{-1})) =(-t^{-1})^n\det(A)\chi_{A^{-1}}(t). $$ That is, $$ \chi_{A^{-1}}(t) = \frac{(-t)^n}{\det(A)}\chi_A(t^{-1}) = t^n + \sum_{i=0}^{n-1}(-1)^{n-i} \frac{c_{n-i}(A)}{c_0(A)} t^{i} $$ But then it follows immediately that $\text{tr}(A^{-1}) = c_{n-1}(A^{-1}) = c_1(A)/c_0(A) = c_1(A)/\det(A)$.

[Note that this expression for $\text{tr}(A^{-1})$ holds for any invertible matrix $A$. The condition that $A$ is symmetric positive definite is not necessary. Indeed we used the fact that $A$ was symmetric because it implies that $A$ is diagonalisable, but this was only used to give an easy proof (in this special case of a symmetric matrix) that $\text{tr}(A)$ is equal to the sum of the eigenvalues of $A$. This can equally be shown by computing the coefficient of $t^{n-1}$ in the characteristic polynomial directly from its definition.]