6.4: Eigenvalue Problems

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We seek nontrivial solutions to the eigenvalue problem

\[A \mathbf{v}=\lambda \mathbf{v} \nonumber \]

We note that \(\mathbf{v}=\mathbf{0}\) is an obvious solution. Furthermore, it does not lead to anything useful. So, it is called a trivial solution. Typically, we are given the matrix \(A\) and have to determine the eigenvalues, \(\lambda\), and the associated eigenvectors, \(\mathbf{v}\), satisfying the above eigenvalue problem. Later in the course we will explore other types of eigenvalue problems.

For now we begin to solve the eigenvalue problem for \(\mathbf{v}=\left(\begin{array}{l}v_{1} \\ v_{2}\end{array}\right)\). Inserting this into Equation \(\PageIndex{1}\), we obtain the homogeneous algebraic system

\[\begin{aligned} &(a-\lambda) v_{1}+b v_{2}=0 \\ &c v_{1}+(d-\lambda) v_{2}=0 \end{aligned} \label{6.68} \]

The solution of such a system would be unique if the determinant of the system is not zero. However, this would give the trivial solution \(v_{1}=0\), \(v_{2}=0 .\) To get a nontrivial solution, we need to force the determinant to be zero. This yields the eigenvalue equation

\[0=\left|\begin{array}{cc} a-\lambda & b \\ c & d-\lambda \end{array}\right|=(a-\lambda)(d-\lambda)-b c \nonumber \]

This is a quadratic equation for the eigenvalues that would lead to nontrivial solutions. If we expand the right side of the equation, we find that

\[\lambda^{2}-(a+d) \lambda+a d-b c=0 \nonumber \]

This is the same equation as the characteristic equation 6.1.12 for the general constant coefficient differential equation considered in the first chapter. Thus, the eigenvalues correspond to the solutions of the characteristic polynomial for the system.

Once we find the eigenvalues, then there are possibly an infinite number solutions to the algebraic system. We will see this in the examples.

So, the process is to

Write the coefficient matrix;
Find the eigenvalues from the equation \(\operatorname{det}(A-\lambda I)=0\); and,
Find the eigenvectors by solving the linear system \((A-\lambda I) \mathbf{v}=0\) for each \(\lambda\).