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 $$A = \begin{bmatrix}
x^2(x – 1)^2 & 0 & 0\\
0 & x(x – 1)(x – 2) & 0 \\
0 & 0 & x(x – 2)^2
\end{bmatrix}$$
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
$$A= \begin{bmatrix}
1 & 0 & 0 & 0 &0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & B \\
\end{bmatrix}$$ for some block matrix $B$.
Best Answer
Can't we just determine the Jordan form from the elementary divisors? $$A = \begin{bmatrix} 0&&&&&&&&&\\ &0&&&&&&&&\\ &&0&&&&&&&\\ &&1&0&&&&&&\\ &&&&1&&&&&\\ &&&&&1&&&&\\ &&&&&1&1&&&\\ &&&&&&&2&&\\ &&&&&&&&2&\\ &&&&&&&&1&2 \end{bmatrix} $$