Triangular Feedback Matrices

An interesting class of feedback matrices, also explored by Jot [217], is that of triangular matrices. A basic fact from linear algebra is that triangular matrices (either lower or upper triangular) have all of their eigenvalues along the diagonal.^4.13 For example, the matrix

$$\mathbf{A}_3 = \left[\begin{array}{ccc} \lambda_1 & 0 & 0\\ [2pt] a & \lambda_2 & 0\\ [2pt] b & c & \lambda_3 \end{array}\right]$$

is lower triangular, and its eigenvalues are $(\lambda_1, \lambda_2,\lambda_3)$ for all values of $a$ , $b$ , and $c$ .

It is important to note that not all triangular matrices are lossless. For example, consider

$$\mathbf{A}_2 = \left[\begin{array}{cc} 1 & 0 \\ [2pt] 1 & 1 \end{array}\right]$$

It has two eigenvalues equal to 1, which looks lossless, but a quick calculation shows that there is only one eigenvector, $[0,1]^T$ . This happens because this matrix is a Jordan block of order 2 corresponding to the repeated eigenvalue $\lambda=1$ . A direct computation shows that

$$\mathbf{A}_2^n = \left[\begin{array}{cc} 1 & 0 \\ [2pt] n & 1 \end{array}\right]$$

which is clearly not lossless.

One way to avoid ``coupled repeated poles'' of this nature is to use non-repeating eigenvalues. Another is to convert $\mathbf{A}$ to Jordan canonical form by means of a similarity transformation, zero any off-diagonal elements, and transform back [332].