Jtdylktuy

I am having trouble undering the relation between invariant factors and it related notions. Could someone help walk me through the following example, which will hopefully clarify things?

Let $T$ be a linear endomorphism on finite-dimensional vector space $V$ over the complex numbers, and let the matrix
$$
left( begin{matrix} x^2(x-1)^2 & 0 & 0 \ 0 & x(x-1)(x-2) & 0 \ 0 & 0 & x(x-2)^2 end{matrix} right).
$$
be the relation matrix of $V$ when views as a $mathbb C [x]$ module under the operation $x cdot alpha = T(alpha )$, for any element $alpha$ in $V$, with respect to the generators ${v_1, v_2, v_3}$.

What are the invariant factors of $T$? Are they just $x^2(x-1)^2$, $x(x-1)(x-2)$, $x(x-2)^2$ or does this matrix need to be adjusted so that the first invariant factor divides the second, et cetera.

What are the elementary divisors of $V$? From this, what is the Jordan form of $T$?

Let $A$ be your matrix. As you suspect, the invariant factors are not just the diagonal entries of $A$. This is clear since they do not satisfy the divisibility condition you mention. However, we can easily obtain the elementary divisors since $A$ is diagonal. As a $mathbb{C}[x]$-module, we have
begin{align*}
V &cong frac{mathbb{C}[x]}{(x^2(x-1)^2)} oplus frac{mathbb{C}[x]}{(x(x-1)(x-2))} oplus frac{mathbb{C}[x]}{(x(x-2)^2)}\
&cong frac{mathbb{C}[x]}{(x^2)} oplus frac{mathbb{C}[x]}{((x-1)^2)} oplus frac{mathbb{C}[x]}{(x)} oplus frac{mathbb{C}[x]}{(x-1)} oplus frac{mathbb{C}[x]}{(x-2)} oplus frac{mathbb{C}[x]}{(x)} oplus frac{mathbb{C}[x]}{((x-2)^2)}
end{align*}
by the Chinese Remainder Theorem. Thus the elementary divisors of $A$ are
$$
x,x,x^2, x-1, (x-1)^2, x-2, (x-2)^2 , .
$$
We can recombine these into the invariant factors $a_1, a_2, a_3$. The largest invariant factor must be divisible by all the elementary divisors, so $a_3 = x^2 (x-1)^2 (x-2)^2$. Looking at what's left, we find that $a_2 = x(x-1)(x-2)$, and finially that $a_1 = x$.

But finding the invariant factors is not actually necessary to compute the Jordan form. From the list of elementary divisors, we see that $T$ has Jordan canonical form
$$
J =
left(begin{array}{r|r|rr|r|rr|r|rr}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &
0 \
hline
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &
0 \
hline
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 &
0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &
0 \
hline
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 &
0 \
hline
0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 &
0 \
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 &
0 \
hline
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 &
0 \
hline
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 &
1 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &
2
end{array}right) , .
$$
See $S12.3$ (p. 491) of Dummit and Foote for more on how to obtain the Jordan form from the list of elementary divisors.