Trying to understand the idea behind $T$-conductor

big-picturegeometryintuitionlinear algebralinear-transformations

I was reading Linear Algebra by Hoffman Kunze and I found a strange definition

Let $W$ be an invariant subspace for $T\in \mathcal L(V)$ and let $\alpha$ be a vector in $V$. The $T$-conductor of $\alpha$ into $W$ is the set $S_T(\alpha; W)$ which consists of all polynomials $g$ (over the scalar field) such that $g(T)\alpha$ is in $W$.

The book also added

In the special case $W=\{0\}$ the conductor is called the $T$-annihilator of $\alpha$.

Now, my problem is I don't understand what's going on here. What's the idea behind these two definitions? What's the algebraic (or geometric) picture behind these ideas? I understand the definiton and the intuition behind all the terms used in defining a $T$-conductor. The problem is I just can't connect them together. Please help me to do so.

Also, just after these two definitions mentioned, they proved the lemma

If $W$ is an invariant subspace for $T$, then $W$ is invariant under every polynomial in $T$. Thus, for each $\alpha$ in $V$, the conductor $S(\alpha; W)$ is an ideal in the polynomial algebra $F[X]$.

and went on to say

The unique monic generator of the ideal $S(\alpha;W)$ is also called the $T$-conductor of $\alpha$ into $W$ (the $T$-annihilator in case $W=\{0\}$).

How come these two definitions are equivalent? This is one more connection I can't get. Please help me here.

It seems that there are a couple of questions on MSE that looks like being similar to this one, but actually, they aren't. In these questions, the OP seems to have an idea of what's going on, whereas I don't. One of these questions even had an example, but that didn't help either.

Best Answer

There are really two questions here. The second (why are these two definitions equivalent?) is easier to answer, so I'll answer that one first.

The lemma states that $S(\alpha; W)$ is an ideal in the polynomial algebra $F[X]$. However, the polynomials form a principal ideal domain, which is to say that we can necessarily say that $$ S(\alpha;W) = \{p(x) f(x): p(x) \in F[X]\} $$ for some polynomial $f(x)$ which is called the "generator" of the ideal. Any polynomial for which the above description is valid is called a generator. Notably, we can always choose $f$ to be monic: for any non-zero $k \in F$, we have $$ p(x)f(x) = (k \cdot p(x)) \cdot \frac{f(x)}{k}, $$ so if we take $k$ to be the leading coefficient of $f(x)$, then the above shows us that $f(x)/k$ is also a generator of the same ideal, and because $k$ is the leading coefficient, $f/k$ will be monic.

Now, it is not exactly correct to say that the definitions of $S(\alpha;W)$ and its unique monic generator (which I will denote as $s(\alpha;W)$) are equivalent. The authors do not say that these are equivalent, only that the same term is used for both of these, which is perhaps an unfortunate example of overloaded terminology in mathematics. That being said, it can be argued that $S(\alpha;W)$ and $s(\alpha;W)$ convey "the same information". Indeed, if we are given the set $S(\alpha;W)$, then $s(\alpha;W)$ is just the lowest-degree monic polynomial in $S(\alpha;W)$. If we are given $s(\alpha;W)$, then $S(\alpha;W)$ is just the set of all polynomials of the form $p(x) \cdot s(\alpha;W)$.

As for the idea behind these definitions: these are useful definitions for the derivation of the cyclic decomposition theorem. The second edition of Hoffmann and Kunze gives the following explanation of how these definitions are used.

Suppose that by some process or another we have selected $\alpha_1,\dots,\alpha_j$ and the subspace $$ W_j = Z(\alpha_1;T) + \cdots + Z(\alpha_j;T) $$ is proper. We would like to find a non-zero vector $\alpha_{j+1}$ such that $$ W_j \cap Z(\alpha_{j+1};T) = \{0\} $$ because the subspace $W_{j+1} = W_j \oplus Z(\alpha_{j+1};T)$ would then come at least one dimension nearer to exhausting $V$. But, why should any such $\alpha_{j+1}$ exist? If $\alpha_1,\dots,\alpha_j$ have been chosen so that $W_j$ is a $T$-admissible subspace, then it is rather easy to see that we can find a suitable $\alpha_{j+1}$. This is what will make our proof of [the cyclic decomposition theorem] work, even if that is not how we phrase the argument.

Let $W$ be a proper $T$-invariant subspace. Let us try to find a non-zero vector $\alpha$ such that $$ W \cap Z(\alpha;T) = \{0\}. $$ We can choose some vector $\beta$ which is not in $W$. Consider the $T$-conductor $S(\beta;W)$, which consists of all polynomials $g$ such that $g(T)\beta$ is in $W$. Recall that the monic polynomial $f = s(\beta;W)$ which generates the ideal $S(\beta;W)$ is also called the $T$-conductor of $\beta$ into $W$. The vector $f(T)\beta$ is in $W$. Now, if $W$ is $T$-admissible, there is a $\gamma$ in $W$ with $f(T)\beta = f(T)\gamma$. Let $\alpha = \beta - \gamma$ and let $g$ be any polynomial. Since $\beta - \alpha$ is in $W$, $g(T)\beta$ will be in $W$ if and only if $g(T)\alpha$ is in $W$; in other words, $S(\alpha;W) = S(\beta;W)$. Thus the polynomial $f$ is also the $T$-conductor of $\alpha$ into $W$. But $f(T)\alpha = 0$. That tells us that $g(T)\alpha$ is in $W$ if and only if $g(T) \alpha = 0$, i.e., the subspaces $Z(\alpha;T)$ and $W$ are independent and $f$ is the $T$-annihilator of $\alpha$.

This is a relatively formal explanation of the importance of $S(\alpha;W)$, so here is a "big picture" perspective. As part of our proof of the cyclic decomposition theorem, we need a convenient way to "verify" whether, for a $T$-invariant subspace $W$ and an arbitrary vector $\alpha$, the invariant subspaces $W$ and $Z(\alpha;T)$ are independent (i.e. $W \cap Z(\alpha;T) = \{0\}$). One nice way to check whether this is the case is to compare the $T$-conductor of $\alpha$ into $W$ with the $T$-annihilator of $\alpha$: these spaces are independent if and only if these polynomials (or equivalently the ideals they generate) are the same. By taking advantage of this, we can start with an arbitrary vector $\beta$, use it to create a new vector $\alpha$, then show that this $\alpha$ has the required property.

To summarize, here is a one-sentence explanation. Together the $T$-annihilator of $\alpha$, the $T$-conductor of $\alpha$ into $W$ describes the (extent of) overlap between $W$ and $Z(\alpha;T)$, the $T$-invariant subspace generated by $\alpha$.

Related Question