I have the following matrix of transfer functions for the linear system $\dot{x}=Ax(t)+Bu(t)$, $y=Cx(t)+Du(t)$. $A$ is a 4x4 matrix, $x$ is a 4x1 vector, $B$ is a 4x2 matrix (not that any of this really matters for the question, but I'm writing it anyways in case there's any confusion as to why there are 4 zero entries in the matrix of transfer functions:
$P(s)=\begin{bmatrix}\frac{s^2+1}{s^4+2s^2-a^2+1}&\frac{-2a}{s^4+2s^2-a^2+1}\\\frac{a}{s^4+2s^2-a^2+1}&\frac{-2(s^2+1)}{s^4+2s^2-a^2+1}\\0&0\\0&0\end{bmatrix}$
The goal is to convert this matrix to Smith-McMillan form. I have attempted the following:
Step 1)
Write $P(s)=\frac{N(s)}{d(s)}$, such that $d(s)=s^4+2s^2-a^2+1$ and $N(s)=\begin{bmatrix}s^2+1&-2a\\a&-2(s^2+1)\\0&0\\0&0\end{bmatrix}$
Step 2)
$\Delta _0=1$
$\Delta _1=gcd(s^2+1,-2a,a,-2(s^2+1))=1$
$\Delta _2=gcd(det(\begin{bmatrix}\frac{s^2+1}{s^4+2s^2-a^2+1}&\frac{-2a}{s^4+2s^2-a^2+1}\\\frac{a}{s^4+2s^2-a^2+1}&\frac{-2(s^2+1)}{s^4+2s^2-a^2+1}\\\end{bmatrix}))=-2s^4-4s^2-2+2a^2$.
Step 3)
$\lambda_1(s)=\frac{\Delta_1}{\Delta_0}=1$
$\lambda_2(s)=\frac{\Delta_2}{\Delta_1}=-2s^4-4s^2-2+2a^2$
Step 4)
Finally, we can find $M(s)=\begin{bmatrix}\frac{\lambda_1}{d(s)}&0\\0&\frac{\lambda_2}{d(s)}\\0&0\\0&0\end{bmatrix}=\begin{bmatrix}\frac{1}{s^4+2s^2-a^2+1}&0\\0&-2\\0&0\\0&0\end{bmatrix}$
Is this correct? It doesn't really look right, just looking at the poles and zeros of $M(s)$, but I'm not sure.