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Are the eigenvalues of the FIR convolution matrix the zeros of the corresponding FIR filter?

Suppose I have an FIR filter $H(z) = h_{0} + h_{1}z^{-1} + h_{2}z^{-2}$. I want to implement it using a matrix, so I have:

$$ H = \begin{bmatrix} h_{0} & 0 & 0 \\ h_{1} & h_{0} & 0 \\ h_{2} & h_{1} & h_{0} \end{bmatrix} $$

Can I find the zeros of $H(z)$ by finding the eigenvalues of $H$?

I know that in general, the roots of a polynomial are given by the eigenvalues of its companion matrix. However, can I just find the eigenvalues of $H$ for this example instead?

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For any convolution matrix (even truncated), your diagonal entries would be $h_0$; hence, the characteristic polynomial, setting $b_0 = h_0-\lambda$,

\begin{align}|H-\lambda I| = \begin{vmatrix} b_0 & 0 & 0\\h_1 & b_0 & 0\\h_2 & h_1 & b_0 \end{vmatrix} &= b_0\,\begin{vmatrix} b_0 & 0\\h_1 & b_0 \end{vmatrix} - 0\,\begin{vmatrix} h_1&0\\h_2&b_0\end{vmatrix} + 0\,\begin{vmatrix} h_1 & b_0\\h_2&h_1 \end{vmatrix} \\ \\ &=b_0\,\begin{vmatrix} b_0 & 0\\h_1 & b_0 \end{vmatrix}\\ &=b_0\left(b_0\,b_0 - 0\,h_1 \right)\\ &=b_0^3 \end{align}

Little induction is necessary to see that for any square matrix of dimension $N\times N$ with an lower left structure and a constant $d$ in each element of the diagonal, the characteristic polynomial is $(d-\lambda)^N$; the Eigenvalues are roots of that polynomial, and hence there's but one Eigenvalue (with $N$ multiplicity): $d$, which in your case is $h_0$.

So, since not all zeros of $H(z)$ are defined by your first filter tap, no, you can't find all zeros of $H(z)$ through Eigenvalues.

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