# Finding Zeros of System Function using Eigenvalues

1. Feb 25, 2014

Hi all - working on this problem wanted to see if anyone had any advice - thanks!

As shown in section 4.4, the poles of the system $H(z)$ with state matrices $\mathbf{A, b, c^t, } d$ are given by the eigenvalues of $\mathbf{A}$.

Find: Show that, if $d\neq0$, the zeros of the system are given by the eigenvalues of the matrix $\left (\mathbf{A}-d^{-1}\mathbf{b}\mathbf{c^t} \right )$.

Hint: The poles of the inverse system $H^{-1}(z)$ equal the zeros of $H(z)$, and $H^{-1}(z)$ has the output $x(n)$ if its input is $y(n)$.

2. $H(z)=\mathbf{c^t}(\mathbf{zI-A})^{-1}\mathbf{b}+d$

3. I understand why the poles of the system are eigenvalues of A. I have gone through this derivation in other work. I feel like there is something I am missing in the linear algebra that would simplify this problem. My attempt at a solution below stops short of solving for eigenvalues of the new matrix because i feel that proving this in generality must be cleaner than this brute force method.

$H^{-1}(z)=\left [\mathbf{c^t}(\mathbf{zI-A})^{-1}\mathbf{b}+d \right ]^{-1}$

$\left ( \mathbf{A}-d^{-1}\mathbf{bc^t} \right ) = \left (\begin{bmatrix} a_{1 1} & \cdots & a_{1 N-1} \\ \vdots & \ddots & \vdots \\ a_{N-1 1} & \cdots & a_{N-1 N-1} \end{bmatrix} - \begin{bmatrix} \frac{b_1c_1}{d} & \cdots & \frac{b1c_{N-1}}{d} \\ \vdots & \ddots & \vdots \\ \frac{b_{N-1}c_1}{d} & \cdots & \frac{b_{N-1}c_{N-1}}{d} \end{bmatrix}\right ) = \begin{bmatrix} a_{1 1} - \frac{b_1c_1}{d} & \cdots & a_{1 N-1} - \frac{b1c_{N-1}}{d} \\ \vdots & \ddots & \vdots \\ a_{N-1 1} - \frac{b_{N-1}c_1}{d} & \cdots & a_{N-1 N-1} - \frac{b_{N-1}c_{N-1}}{d} \end{bmatrix} = \mathbf{A'}$

And then some eigen decomposition leads towards....

$|\mathbf{A'}-\lambda\mathbf{I}|=0 = det\begin{bmatrix} a_{1 1} - \frac{b_1c_1}{d}-\lambda & \cdots & a_{1 N-1} - \frac{b1c_{N-1}}{d}-\lambda \\ \vdots & \ddots & \vdots \\ a_{N-1 1} - \frac{b_{N-1}c_1}{d} -\lambda & \cdots & a_{N-1 N-1} - \frac{b_{N-1}c_{N-1}}{d}-\lambda \end{bmatrix}$

Is there something in the composition of $\mathbf{A,b,c^t,} d$ that I am missing?

Thanks all.

Last edited: Feb 25, 2014
2. Feb 25, 2014

### donpacino

Just using the equation is the cleanest method I know of.
if you use matlab this method is extremely easy to implement.
I just want to note that you made an error in your step #3

Last edited: Feb 25, 2014
3. Feb 25, 2014

Thank you for your response donpacino.

Unfortunately I'm not seeing it. (both the way to implement the equation and the error in 3)
I considered MATLAB but I was hesitant to take a deterministic approach and go with "if it works this once it will always work" kind of deal. Maybe i'l do that.

4. Feb 26, 2014

### donpacino

I= the identity matrix, which means the diagonal is equal to 1, and zero everywhere else

I=
[1 0 0
0 1 0
0 0 1]

so if A=
[1 1 1
1 1 1
1 1 1]

then |A-λI|=
[1-λ, 1, 1;
1, 1-λ , 1;
1, 1, 1-λ]

does that make sense?

Last edited: Feb 26, 2014
5. Feb 27, 2014