In order to find the eigenvalues of a square matrix $A$, we must find the values of $\lambda$ such that the equation \begin{equation} A \left|v\right> = \lambda \left|v\right> \end{equation} admits solutions $\left|v\right>$. (The solutions $\left|v\right>$ are eigenvectors of $A$, as discussed in the next section.) Rearranging terms, $\left|v\right>$ must satisfy \begin{equation} (\lambda I-A)\left|v\right> = 0 \label{eeq} \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 (\ref{eeq}) by this inverse would then yield \begin{equation} (\lambda I-A)^{-1}(\lambda I-A)\left|v\right> = \left|v\right> = 0 \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 $\det A=0$.

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

Example

Suppose $A=\begin{pmatrix}1&2\\9&4\\\end{pmatrix}$. Then we must solve \begin{equation} 0 = |\lambda I-A| = \begin{vmatrix}1&2\\9&4\\\end{vmatrix} = (\lambda-1)(\lambda-4)-18 \end{equation} or equivalently \begin{equation} 0 = \lambda^2 - 5\lambda -14 = (\lambda-7)(\lambda+2) , \end{equation} and the eigenvalues in this case are $\lambda=7$ and $\lambda=-2$.

In general, this computation always yields a polynomial equation in $\lambda$, 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!