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Physics Linear Algebra I Studied #22

Is Further Block Decomposition Always Possible?

Poking at a special case where the minimal and characteristic polynomials match as a power of an irreducible polynomial, and what that tells us about block decomposition.

Written · 5 min read·By gdparkEnglish edition updated

Today we’re gonna look at some special cases and poke at their properties. But — if this is the first post of mine you’re seeing, this is gonna feel like it dropped out of nowhere, so let me just paste in the conclusion from the very end of the last post. Where did we leave off last time —

Summary table from the last post: m_A,X(t), m_A(t) and φ_A(t) with their names, definitions and degrees k_X ≤ k ≤ n

Right here. That’s as far as we’d sorted things out.

A special case

OK so — what special case are we gonna look at?

What if k_X = n is the case?!?! What happens then??

If that holds, the inequality that used to be k_X ≤ k ≤ n collapses into k_X = k = n, like this… Which means all three polynomials have the same degree, and since each one divides the next (and they’re all monic),

All three polynomials equal: m_A,X(t) = m_A(t) = φ_A(t)

it has to come out like this. And what this is telling us is — among all the matrices similar to A, there’s an $A'$ that can be written as a companion matrix.

Being written as a companion matrix meant that the basis that builds $A'$ has the shape

The basis β′ = {X, AX, …, Aⁿ⁻¹X}

like this, so it must mean there’s an X that, satisfying m_A,X(t) = φ_A(t), makes

The basis β′ = {X, AX, …, Aⁿ⁻¹X}

possible, right?!?!

Yep. That’s the one. There’s an X that makes it happen!!! When k = n, that is?!?! What this is really saying is: it’s the same as asking what happens when m_A(t) = φ_A(t). And what on earth does that even mean…… It’s not hard!!!

The special case: m_A(t) = φ_A(t) = p^m(t), with p(t) irreducible

Here, $p(t)$ is an irreducible polynomial. Let’s say the minimal polynomial and the characteristic polynomial are the same — but they’re the same in the form of a power of an irreducible polynomial. We set up the situation above because, starting from a special case like this, we can talk about general properties.

The chain ker I ≤ ker p(A) ≤ ··· ≤ ker p(A)^(m−1) ≤ ker p(A)^m

$\ker B$ is a set of vectors — specifically, the set of vectors X satisfying $BX = 0$. So X ∈ ker p(A)^ℓ means X is one of the X’s satisfying p(A)^ℓ·X = 0. And if we multiply both sides of that equation by $p(A)$, p(A)^(ℓ+1)·X = 0 holds automatically too. So the chain of inclusions up there holds automatically, of course.

Look at the far left end. X belonging to $\ker I$ means X satisfies $IX = 0$ — so the only thing that fits is the zero vector. ∴ $\ker I = \{0\}$ and the dimension is 0…

On the far right, the X’s belonging to ker p(A)^m are the X’s satisfying p(A)^m X = 0.

But is there even an X that satisfies this?!?!

→ There are a TON of them lol!!! It holds for absolutely every single vector!!! Because!!! p(A)^m = φ_A(A), so it’s just 0!!! So it holds for all vectors.

∴ $\ker p(A)^m$ is a vector space containing every vector in sight!!! Its dimension is $n$!!!

If you draw this as a Venn diagram, apparently it looks like this.

Venn diagram: ker I with just 0 in it, inside ker p(A)^(m−1), inside ker p(A)^m

A proposition

Now I’m gonna toss out a proposition.

Proposition: there is an X not in ker p(A)^(m−1), and for that X, m_A,X(t) = p(t)^m

Here, in an X not in ker p(A)^(m−1), X does not belong to $\ker p(A)^{m-1}$, so we’re talking about a vector X that does not land inside $\ker p(A)^{m-1}$ but lives only inside $\ker p(A)^m$. And for that X, m_A,X(t) ≠ p(t)^(m−1) holds, and m_A,X(t) = p(t)^m is the case!!!

Now — b.u.t. !!!

Outside ker p(A)^(m−1), does an X like that — one that only lives inside ker p(A)^m — actually exist?!?!

Yep….. it exists with 100% probability lol!! Because what we assumed means ker p(A)^(m−1) and ker p(A)^m can never, ever be equal, not in a million years. If the two sets were equal, p(A)^(m−1)X = 0 would hold for every X, so the minimal polynomial of A would become p(t)^(m−1) (or something that divides it). And then the assumption we set up at the very start would be broken. Can’t allow that.

That is,

The same Venn diagram with a red star in the ring between ker p(A)^(m−1) and ker p(A)^m

an element like the red star always!!! exists!!!

That is,

If k = n, then automatically k_X = n for some X

this is one property!!! Yeah!

Another important property

A matrix A like this one — with k = n, and with everything a power of a single irreducible $p(t)$ — has yet another important property. And that one is~~~~~

The block form A′ = (B, 0; 0, C)

a similar matrix $A'$ decomposed like this can absolutely never be made.

Let me prove it. The assumption is “decomposition is possible” — I’ll assume that and roll with it.

Sizes: B is ℓ × ℓ and C is (n − ℓ) × (n − ℓ)

I’ll assume the sizes look like this, and φ_A(t) = φ_B(t)φ_C(t) — that part is of course obvious. And here, we said φ_A(t) = p(t)^m, right?!?! Then φ_B(t) and φ_C(t) will each be written as

Each one a power of p: φ_B(t) = p(t)^s and φ_C(t) = p(t)^(m−s)

like this.

OK so, deg of φ_B(t) = ℓ, but that would have to be (deg of p(t))·s, wouldn’t it?!?! By the same logic, deg of φ_C(t) = n − ℓ = (deg of p(t))·(m − s), riiight~~~~

Also, m_B(t) is a divisor of φ_B(t), and m_C(t) is a divisor of φ_C(t). The ingredients are all on the counter now.

Let’s think about m_A(t). Well, m_A(t) would be lcm(m_B(t), m_C(t)), and if we work it out using the ingredients we just prepped,

Least common multiple: lcm(m_B(t), m_C(t)) = p(t)^q, where q is less than m

the conclusion that pops out is — $q$ has to be strictly less than $m$.

Whoa~~~~~~ but we said m_A(t) = p(t)^m!!! That is, the result we got is that they can absolutely never be equal — so it’s a contradiction!

What was the contradiction?!?!

“Assuming decomposition is possible” — that was the contradiction.

(In the situation where φ_A = p(t)^m = m_A(t) holds, I mean.) Q.E.D. — quod erat demonstrandum.

That was the moment one of those long-simmered, long-simmered questions finally got resolved!!!

Our very first question was “Is diagonalization possible for all matrices?!?!”

↓

“Ohhhhh so if we just keep blocking and blocking, it’ll end up diagonalized?!?!”

↓

“Then is there a case where blocking is not possible?!?!”

↓

And now we get to add one more piece of flesh to the bone right here.

Blocking A step by step until it becomes block-diagonal with blocks A₁, …, A_k

Handwritten: A₁ to A_k all reach φ_Aᵢ(t) = m_Aᵢ(t) = Pᵢ(t)^(mᵢ); the blocking goes on until then, and from there no more blocking is possible

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

  • The proposition is about an $X$ outside the smaller kernel: $\exists X \notin \ker p(A)^{m-1}$, and for that $X$, $m_{A,X}(t) = p(t)^m$ (the pictures write $\in$; for an $X$ in $\ker p(A)^{m-1}$, $m_{A,X}(t)$ divides $p(t)^{m-1}$).
  • In the least common multiple, $m_B(t)$ only divides $p(t)^s$ and $m_C(t)$ only divides $p(t)^{m-s}$, so $\operatorname{lcm}(m_B(t), m_C(t))$ divides $\operatorname{lcm}(p(t)^s, p(t)^{m-s})$ rather than equalling it; either way it is $p(t)^q$ with $q \le \max(s, m - s) \lt m$, so the conclusion is right.

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