Let \(V\) be an \(n\)-dimensional vector space over \(F\). What is the characteristic polynomial of the identity operator on \(V\)? What is the characteristic polynomial for the zero operator?
Solution
If \(I_V\) is the identity operator, then in any ordered basis its matrix is \(I_n\). Hence \[ \chi_{I_V}(x)=\det(xI_n-I_n)=\det((x-1)I_n)=(x-1)^n. \] For the zero operator, the matrix is \(0\), so \[ \chi_0(x)=\det(xI_n)=x^n. \]