Method for Finding Eigenvalues of a Square Matrix.
In order to find the eigenvalues of a square matrix \(A\text{,}\) we must find the values of \(\lambda\) such that the equation
\begin{equation}
A \left|v\right> = \lambda \left|v\right>\tag{4.2.1}
\end{equation}
admits solutions \(\left|v\right>\text{.}\) (The solutions \(\left|v\right>\) are eigenvectors of \(A\text{,}\) as discussed in the next section.) Rearranging terms, \(\left|v\right>\) must satisfy
where \(I\) denotes the identity matrix (of the same size as \(A\)). Suppose that the inverse matrix \((\lambda I-A)^{-1}\) exists. Multiplying both sides of (4.2.2) by this inverse would then yield
which is not a very interesting solution. So, instead, we want \((\lambda I-A)\)not to have an inverse. When does this happen?
Claim: A square matrix \(A\) does not have an inverse if and only if \(\det A=0\text{.}\)
Theorem4.2.The Eigenvalues of a Square Matrix \(\boldsymbol{A}\).
Thus, the eigenvalues of \(A\) are the solutions of the equation \(\det(\lambda I-A)=0\text{.}\)Usually, this computation yields a polynomial equation in \(\lambda\text{,}\) which is of order \(n\) if \(A\) is an \(n\times n\) matrix. Thus, we expect \(n\) (complex) eigenvalues, which however might not be distinct.
Definition4.3.Degeneracy/Multiplicity.
In the case of an eigenvalue that is repeated \(m\) times, we call the eigenvalue \(\boldsymbol{m}\)-fold degenerate, or, equivalently, of multiplicity \(\boldsymbol{m}\). For more information about degenerate eigenvalues, see Section 4.6.
Example.
Suppose \(A=\begin{pmatrix}1\amp 2\\9\amp 4
\end{pmatrix}\text{.}\) Then we must solve
