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Section 4.2 Finding Eigenvalues

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

\begin{equation} (\lambda I-A)\left|v\right> = 0\tag{4.2.2} \end{equation}

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

\begin{equation} (\lambda I-A)^{-1}(\lambda I-A)\left|v\right> = \left|v\right> = 0\tag{4.2.3} \end{equation}

which is not a very interesting solution. So 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 \(\det A=0\text{.}\)

Thus, the eigenvalues of \(A\) are the solutions of the equation \(\det(\lambda I-A)=0\text{.}\)


Suppose \(A=\begin{pmatrix}1\amp 2\\9\amp 4 \end{pmatrix}\text{.}\) Then we must solve

\begin{equation} 0 = |\lambda I-A| = \begin{vmatrix}1\amp 2\\9\amp 4 \end{vmatrix} = (\lambda-1)(\lambda-4)-18\tag{4.2.4} \end{equation}

or equivalently

\begin{equation} 0 = \lambda^2 - 5\lambda -14 = (\lambda-7)(\lambda+2)\text{,}\tag{4.2.5} \end{equation}

and the eigenvalues in this case are \(\lambda=7\) and \(\lambda=-2\text{.}\)

In general, this computation always 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. In the case of an eigenvalue that is repeated \(m\) times, we call the eigenvalue \(m\)-fold degenerate.