8.1: Introduction to the Eigenvalue Problem
- Page ID
- 21847
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Harking back to our previous discussion of The Laplace Transform we labeled the complex number \(\lambda\) an eigenvalue of \(B\) if \(\lambda I-B\) was not invertible. In order to find such \(\lambda\) one has only to find those \(s\) for which \((sI-B)^{-1}\) is not defined. To take a concrete example we note that if
\[B = \begin{pmatrix} {1}&{0}&{0}\\ {1}&{1}&{0}\\ {0}&{0}&{2} \end{pmatrix} \nonumber\]
then
\[(sI-B)^{-1} = \frac{1}{(s-1)^{2} (s-2)} \begin{pmatrix} {(s-1)(s-2)}&{s-2}&{0}\\ {0}&{(s-1)(s-2)}&{0}\\ {0}&{0}&{(s-1)^2} \end{pmatrix} \nonumber\]
and so \(\lambda_{1} = 1\) and \(\lambda_{2} = 2\) are the two eigenvalues of \(B\). Now, to say that \(\lambda_{j}I-B\) is not invertible is to say that its columns are linearly dependent, or, equivalently, that the null space \(\mathscr{N} (\lambda_{j}I-B)\) contains more than just the zero vector. We call \(\mathscr{N} (\lambda_{j}I-B)\) the jth eigenspace and call each of its nonzero members a jth eigenvector. The dimension of \(\mathscr{N} (\lambda_{j}I-B)\) is referred to as the geometric multiplicity of \(\lambda_{j}\). With respect to \(B\) above, we compute \(\mathscr{N} (\lambda_{j}I-B)\) by solving \((I-B)x = 0\) i.e.,
\[\begin{pmatrix} {0}&{-1}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{1} \end{pmatrix} \begin{pmatrix} {x_{1}}\\ {x_{2}}\\ {x_{3}} \end{pmatrix} = \begin{pmatrix} {0}\\ {0}\\ {0} \end{pmatrix} \nonumber\]
Clearly,
\[\mathscr{N} (\lambda_{1}I-B) = \{c \begin{pmatrix} {1}&{0}&{0} \end{pmatrix}^{T} | c \in \mathbb{R}\} \nonumber\]
Arguing along the same lines we also find
\[\mathscr{N} (\lambda_{2}I-B) = \{c \begin{pmatrix} {0}&{0}&{1} \end{pmatrix}^{T} | c \in \mathbb{R}\} \nonumber\]
That \(B\) is \(3 \times 3\) but possesses only 2 linearly eigenvectors leads us to speak of \(B\) as defective. The cause of its defect is most likely the fact that \(\lambda_{1}\) is a double pole of \((sI-B)^{-1}\). In order to flesh out that remark and uncover the missing eigenvector we must take a much closer look at the transfer function
\[R(s) \equiv (sI-B)^{-1} \nonumber\]
In the mathematical literature this quantity is typically referred to as the Resolvent of \(B\).