I want to prove the following theorem
Let $V$ be a finite dimensional vector space over $\mathbb{R}$ and $f(x)$ be the characteristic polynomial of a linear operator $T: V \rightarrow V$. Then $f(0) \neq 0$ iff $T$ is invertible.
It seems this question is related with my previous questions $f(A)$ is invertible iff characteristic polynmoial and f have no common root
I think I can change $T$ to be matrix $A$, and set characteristic polynomial $f(t) = det(A – t I_n)$, then clearly $f(0) = det(A) $ so $det(A)\neq 0$ iff $A$ is invertible.
Is this the right process? Without introducing $A$, how one can show this?
Best Answer
For any polynomial $f$ with leading coefficient $1$ the constant term $f(0)$ is the product of the roots. In this case the $f(0) \neq 0$ iff the product of the eigen values is not $0$ iff $0$ is not an eigen value off $T$ is invertible,