Inverse of a symmetric using Cayley-Hamilton theorem

Question

Let $A$ be a symmetric positive definite matrix. I am interested in finding a general expression for the $A^{-1}$. Using the Cayley-Hamilton Theorem (https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem), we have that $A^{-1}$ can be expressed as \begin{aligned}A^{-1}={\frac {(-1)^{n-1}}{\det A}}(A^{n-1}+c_{n-1}A^{n-2}+\cdots +c_{1}I_{n}),\end{aligned}

I am bit puzzled finding the coefficients $c_i$, when $A$ is symmetric and positive definite. Also, would it be possible to avoid evaluating the $det{A}$, that is finding a expression equivalent to it?

Answer 1

$$\det(\lambda I-A)=\lambda^n+\beta_1\lambda^{n-1}+...+\beta_n,$$ where $$\beta_i=(-1)^i( \text { the sum of the principal minors of A of order }i),1 \le i \le n.$$ Note that $\beta_n=(-1)^n\det(A).$The easiest way to evaluate a determinant, whether $\det(A)$or any other of the principal minors of $A$ is by pivotal condensation, which is automatic row-reduction.However, there is an alternate formula, due to Bocher, for evaluating the $\beta_i$ $$\beta_1=-\text {trace}(A),$$ $$\beta_i=\frac{-1}{i}[1,\beta_1,...,\beta_{i-1}]\bullet [\mu_i,\mu_{i-1},...,\mu_1],2 \le i \le n,$$ where $\mu_i=$trace$(A^i),1 \le i \le n.$ Note that the most efficient way to calculate such an expression as $$A^3+\beta_1 A^2+\beta_2A+\beta_3I$$ is by "synthetic division" $$A(A(A+\beta_1I)+\beta_2I)+\beta_3I$$ working outwards from the innermost parenthesis and calculating successive powers of $A$, the $\mu_i$ and the $\beta_i$ recursively. Bocher's formula is valid over any field of characteristic 0 or of characteristic > $n.$