## Section5.2Finding 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{5.2.1} \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\label{eeq}\tag{5.2.2} \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 (5.2.2) by this inverse would then yield

\begin{equation} (\lambda I-A)^{-1}(\lambda I-A)\left|v\right> = \left|v\right> = 0\tag{5.2.3} \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 $\det(\lambda I-A)=0\text{.}$

Example

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{5.2.4} \end{equation}

or equivalently

\begin{equation} 0 = \lambda^2 - 5\lambda -14 = (\lambda-7)(\lambda+2)\text{,}\tag{5.2.5} \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 $n$ eigenvalues, which however might not be distinct. Furthermore, not all of the eigenvalues will be real!