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## Section4.2Finding Eigenvalues

In order to find the eigenvalues of a square matrix $$A\text{,}$$ we must find the values of $$\lambda$$ such that the equation

$$A \left|v\right> = \lambda \left|v\right>\tag{4.2.1}$$

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

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

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 ofÂ (4.2.2) by this inverse would then yield

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

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

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

or equivalently

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

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.