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Triangular Feedback Matrices

An interesting class of feedback matrices, also explored by Jot [219], 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.3.13 For example, the matrix

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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 [336].


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written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Mu