Find the invariant factors and elementary divisors from the relations matrix. Let $V$ be a finite dimensional vector space over $\Bbb C$, and $T$ be a linear operator on $V$. Consider $V$ as an $\Bbb C[x]$-module by defining $xv = T(v)$ for each $v \in V$.
Let $$A = \begin{bmatrix}
    x^2(x - 1)^2      &amp; 0 &amp; 0\\
   0       &amp; x(x - 1)(x - 2) &amp; 0 \\
   0      &amp; 0 &amp; x(x - 2)^2
\end{bmatrix}$$
be the relation matrix for $V$ with respect to $\{v_1, v_2, v_3\}$.

List the invariant factors and elementary divisors of $V$.

Attempt: In general, I would reduce the relation matrix to smith normal form to find the invariant factors, and from there I could find the elementary divisors. However, this matrix is hard to reduce to SNF, so I think they want me to find another approach.
From the relation matrix I believe $$V = \frac{F[x]}{x^2(x - 1)^2} \oplus \frac{F[x]}{x(x - 1)(x-2)} \oplus \frac{F[x]}{x(x - 1)^2}$$ and the elementary divisors are $ x^2, (x-2)^2, x, x, (x-1)^2, x - 1, x - 2$, which would mean the cyclic decomposition is
$$V=\frac{F[x]}{x^2(x - 1)^2(x-2)^2} \oplus \frac{F[x]}{x(x - 1)(x-2)} \oplus \frac{F[x]}{x}$$ with invariant factors $x^2(x - 1)^2(x-2)^2, x(x - 1)(x-2), x$.

Describe the decomposition of $V$ by the primary decomposition theorem, and find the Jordan form of the operator $T$.

Attempt: From the invariant factors, I believe the minimal polynomial $p = x^2(x - 1)^2(x-2)^2$, and the characteristic polynomial is $ f = x^2(x - 1)^2(x-2)^2x(x - 1)(x-2)x$. Hence if $N_1 = \mathrm{null}(T^2), N_2 = \mathrm{null}(T - I)^2, N_3 = \mathrm{null}(T- 2I)^2$, then $V=N_1 \oplus N_2 \oplus N_3$ is the primary decomposition. I'm not sure, but I think the Jordan form will be
$$A= \begin{bmatrix}
   1      &amp; 0 &amp; 0 &amp; 0 &amp;0\\
   0       &amp; 1 &amp; 0 &amp; 0 &amp; 0\\
   0      &amp; 0 &amp; 2 &amp; 0 &amp; 0 \\
   0      &amp; 0 &amp; 0 &amp; 2 &amp; 0 \\
   0      &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
   0      &amp; 0 &amp; 0 &amp; 0 &amp; B \\
\end{bmatrix}$$ for some block matrix $B$.
Can't we just determine the Jordan form from the elementary divisors? 
$$A = \begin{bmatrix}
0&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;\\
&amp;0&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;\\
&amp;&amp;0&amp;&amp;&amp;&amp;&amp;&amp;&amp;\\
&amp;&amp;1&amp;0&amp;&amp;&amp;&amp;&amp;&amp;\\
&amp;&amp;&amp;&amp;1&amp;&amp;&amp;&amp;&amp;\\
&amp;&amp;&amp;&amp;&amp;1&amp;&amp;&amp;&amp;\\
&amp;&amp;&amp;&amp;&amp;1&amp;1&amp;&amp;&amp;\\
&amp;&amp;&amp;&amp;&amp;&amp;&amp;2&amp;&amp;\\
&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;2&amp;\\
&amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp;1&amp;2
\end{bmatrix}
$$
1) Your answer to this question is right (but a comma is missing in the list of elementary divisors).

2) From the list of elementary divisors you have written the answer is the matrix given by user346069