Table of contents
Definition
EXAMPLE 1. The vector \(\mathbf{v}=\begin{bmatrix}1\\1\end{bmatrix}\) is an eigenvector of eigenvalue \(-2\) for the matrix \(A=\begin{bmatrix} -1&3\\3&-1\end{bmatrix}\).
Indeed, if we multiply the matrix and the vector we obtain
\[A\mathbf{v} = \begin{bmatrix} -1&3\\3&-1\end{bmatrix} \cdot \begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}2\\2\end{bmatrix} = 2 \begin{bmatrix}1\\1\end{bmatrix} = 2\mathbf{v}\]Finding Egienvalues and Eigenvectors
Eigenvectors
We can observe that \(A\mathbf{x}=\lambda\mathbf{x}\) is equivalent to
\[\begin{equation} (A-\lambda I)\mathbf{x} = \mathbf{0}, \label{eq:eigs} \end{equation}\]where \(I\) is the identity matrix. Indeed, \(\lambda I \mathbf{x} =\lambda \mathbf{x}\).
Therefore, finding the eigenvalues and associated eigenvectors of a matrix is equivalent to finding nontrivial solutions to the system of equations described in \eqref{eq:eigs}.
NOTE: While a vector solution \(\mathbf{x}\) cannot be zero, the corresponding eigenvalue \(\lambda\) could be zero.
EXAMPLE 2. The matrix \(A=\begin{bmatrix} -2 & 3 & -2 \\ 1 & 0 & -2 \\ 1 & 3 & -5 \end{bmatrix}\) has \(-3\) as one of its eigenvalues. Let’s find the corresponding eigenspace. First, we calculate \(A-(-3)I\),
\[A-(-3)I = A+3I = \begin{bmatrix} 1 & 3 & -2 \\ 1 & 3 & -2 \\ 1 & 3 & -2 \end{bmatrix}.\]Performing basic row operations (R2-1\(\times\)R1) and (R3-1\(\times\)R1) we obtain the matrix
\[\begin{bmatrix}1 & 3 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}.\]Now, considering \((A-\lambda I)\mathbf{x}=\mathbf{0}\), and \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\) we obtain the equation \begin{equation} x_1+3x_2-2x_3=0. \label{eq:eigvec_Ex2} \end{equation}
Isolating \(x_1\) in \eqref{eq:eigvec_Ex2}, we obtain
\[\mathbf{x} = \begin{bmatrix} -3x_2+2x_3 \\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix}-3x_2\\x_2\\0\end{bmatrix} + \begin{bmatrix}2x_3\\0\\x_3\end{bmatrix}=x_2\begin{bmatrix}-3\\1\\0\end{bmatrix} + x_3\begin{bmatrix}2\\0\\1\end{bmatrix}\]with \(x_2\) and \(x_3\) free variables. So the eigenspace for \(\lambda=-3\) is a subspace of dimension \(2\) in \(\mathbb{R}^3\) with basis
\[\left\lbrace \begin{bmatrix}-3\\1\\0\end{bmatrix}, \begin{bmatrix}2\\0\\1\end{bmatrix} \right\rbrace\]The method used in the previous example provides a general method for finding eigenvectors and eigenspaces when the eigenvalues are known. But, how do we find eigenvalues?
Eigenvalues
To find the eigenvalues, we need to find the nontrivial solutions to equation \eqref{eq:eigs}. So we need to determine those values of \(\lambda\) for which equation \eqref{eq:eigs} has a nontrivial solution.
This problem is equivalent to finding those values of \(\lambda\) for which the matrix \((A-\lambda I)\) is singular (or its inverse does not exist). That is, we need to find those values of \(\lambda\) such that
\begin{equation} \text{det}(A-\lambda I)=0. \label{eq:charac_eq} \end{equation}
This scalar equation is called characteristic equation.
EXAMPLE 3. In the previous example, we saw that the matrix
\[A=\begin{bmatrix} -2 & 3 & -2 \\ 1 & 0 & -2 \\ 1 & 3 & -5 \end{bmatrix}\]has \(-3\) as one of its eigenvalues. But what are all the eigenvalues of \(A\)?
We will use the characteristic equation \eqref{eq:charac_eq}.
First we compute \(A-\lambda I\):
\[A-\lambda I = \begin{bmatrix} -{\lambda} - 2 & 3 & -2 \\ 1 & -{\lambda} & -2 \\ 1 & 3 & -{\lambda} - 5\end{bmatrix},\]and then we find the determinant:
\[\text{det}(A-\lambda I) = -{\lambda}^{3} - 7 \, {\lambda}^{2} - 15 \, {\lambda} - 9\]Since we know that \(-3\) is a zero of the equation above, we can factor \(\lambda+3\):
\[\begin{align*} \text{det}(A-\lambda I) &= (\lambda+3)(-\lambda^2 - 4\lambda - 3) \\ &= -(\lambda+3)^2(\lambda+1) \end{align*}\]So we can see that the eigenvalues are \(\lambda=-3\) and \(\lambda=-1\).