Section 6.4 The number of eigenvalues of a matrix of order \(n\)
If \(A\) is a matrix of order \(n\text{,}\) then \(p(\lambda)=\det(A-\lambda I)\) is a polynomial of degree \(n\text{.}\) Since a polynomial of degree \(n\) has at most \(n\) roots, the matrix has as most \(n\) eigenvalues.
Theorem 6.4.1. The number of eigenvalues of \(A\).
A square matrix of order \(n\) has at most \(n\) eigenvalues.
Proof.
The characteristic polynomial \(p(\lambda)\) is of degree \(n\text{,}\) and such a polynomial has at most \(n\) real roots.
If we once again look at the matrix from Example 6.1.6. we see that it is a matrix of order \(4\) with \(4\) eigenvalues given. Theorem 6.4.1 tells us that there are no others.