Is there a polynomial $f$ such that $f(A)=I$?

Question

Suppose that $A\in M_n(\mathbb{F})$ and $A$ is nonsingular. Is there a polynomial $f\in \mathbb{F}[x]$ such that $f(A)=I$?

I am thinking of the constant polynomial $f(x)=1$. Am I correct with this one? Are there really such polynomials?

Answer 1

Consider the characteristic polynomial $p_A(\lambda) = \det(A - \lambda I)$.

Let $p_A(\lambda) = a_0+ a_1\lambda + \ldots + a_n\lambda^n$.

We have $$0 \ne \det(A) = p_A(0)= a_0$$

On the other hand, Cayley-Hamilton states that $$0 = p_A(A) = a_0I+ a_1A+ \ldots + a_nA^n = a_0I + A(a_1 + a_2A + \cdots + a_nA^{n-1})$$

Rearranging gives

$$-\frac{1}{a_0}A(a_1+ a_2A +\cdots + a_nA^{n-1}) = I$$

so the desired polynomial is $$f(x) = -\frac{x}{a_0}(a_1 + a_2x + \cdots + a_nx^{n-1}) = -\frac{a_1}{a_0}x -\frac{a_2}{a_0}x^2 -\cdots -\frac{a_n}{a_0}x^n$$

Answer 2

For a non trivial one, Consider the minimal polynomial $f$ of $A$, $f=a_nX^n+...+a_0$, $a_0\neq o$, $g=-(f-a_0)/a_0$.