Show that E_n - λN is invertible for every λ∈K and determine (E_n - λN)^-1

Question

Could you guys help me with this task? Let K be a field and N ∈ Mat (n×n, K) with N^k+1 = 0 for a k∈N. Show that E_n - λN is invertible for every λ∈K and determine (E_n - λN)^-1. What does this mean for the eigenvalues of N?

Thanks in advance!

Answer 1

It is $$(E_n - \lambda N)(E_n + \lambda N + \dots + \lambda^k N^k) = E_n- \lambda^{k+1} N^{k+1} = E_n,$$ meaning that $E_n - \lambda N$ is invertible and its inverse is $E_n + \lambda N + \dots + \lambda^k N^k$.

(If you want, I can explain this answer in German as well.)

Answer 2

Hint: Consider the product $$ (E_n - \lambda N)(E_n + \lambda N + \lambda^2 N^2 + \cdots + \lambda^k N^k) $$ Note that if $(E_n - \lambda N)$ is invertible and $\lambda \neq 0$, then $(\lambda^{-1} E_n - N)$ is invertible. This implies that $\lambda^{-1}$ is not an eigenvalue of $N$.

Answer 3

1. A nilpotent matrix has only one eigenvalue: $0$.

2. For $ \lambda=0$ we have $E_n - λN=E_n$, hence $E_n - λN$ is invertible.

3. Let $ \lambda \ne 0$. Then $E_n - λN$ is invertible $ \iff N- \frac{1}{\lambda}E_n$ is invertible $ \iff \frac{1}{\lambda}$ is not an eigenvalue of $N$. Now look at 1.