prove that n is an eigenvalue of an n by n matrix A, and find a corresponding eigenvector

Question

I have an n by n matrix. How do I prove that n is an eigenvalue of this matrix? I know it will have at most n distinct eigenvalues, but I'm not sure how to proceed from there...

Also, how do I then find its corresponding eigenvector?

Answer 1

HINT

By definition $n$ is an eigenvalue if

• $A\vec v=n \vec v \iff (A-nI)\vec v=0$ for $\vec v\neq 0 \iff det(A-nI)=0$
• once we have found $n$ then from $(A-nI)\vec v=0$ find the eigenvector(s) $\vec v$