Introduction Harking back to our previous discussion of The Laplace Transform we labeled the complex number λ an eigenvalue of B if λ I B was not invertible. In order to find such λ one has only to find those s for which s I B is not defined. To take a concrete example we note that if B 110 010 002 then s I B 1 s 1 2 s 2 s1 s2 s2 0 0 s1 s2 0 0 0 s1 2 and so λ1 1 and λ2 2 are the two eigenvalues of B. Now, to say that λj I B is not invertible is to say that its columns are linearly dependent, or, equivalently, that the null space 𝒩 λj I B contains more than just the zero vector. We call 𝒩 λj I B the jth eigenspace and call each of its nonzero members a jth eigenvector. The dimension of 𝒩 λj I B is referred to as the geometric multiplicity of λj . With respect to B above, we compute 𝒩 λ1 I B by solving IB x 0 , i.e., 0 -1 0 000 001 x1 x2 x3 0 0 0 Clearly 𝒩 λ1 I B c 100 c Arguing along the same lines we also find 𝒩 λ2 I B c 001 c That B is 3x3 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 λ1 is a double pole of sI B 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 sI B In the mathematical literature this quantity is typically referred to as the Resolvent of B.