$[Linear Algebra] $T$-invariant subspace$

Question

$[Linear Algebra] $T$-invariant subspace$

Please see attached image.
Advanced Linear Algebra exercise. Class is working out of Hoffman/Kunze Linear Algebra.
Any questions about notation or what theorems we have acess to, just let me know.

Some definitions:
(1) A linear operator $T$ over a vector space $V$ is diagonalizable if $V$ has a basis whose elements are eigenvectors of $T$. Alternatively, $T$ is diagonalizable if its eigenvectors generate $V$.
(2) Let $T$ denote a linear operator over a vector space $V$. A subspace $W$ of $V$ is $T$-invariant if $T(W) ⊆ W$.
(3) If $W$ $\leq $ $V$ is $T$-invariant (with $T$ $\in $ $L(V)$) the compression of $T$ to $W$ is $Tw:=$ $T|_{W}^{W} $ $\in $ $L(W)$.

Linear Algebra Vector Spaces Eigenvalues & Eigenvectors

Matchmaticians Alumnus

211

Matchmaticians [Linear Algebra] $T$-invariant subspace File #1

File #1 (png)

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

Let $\{e_i \mid i \in I\}$ be a basis of eigenvectors of $T$, which exists by hypothesis. For $J \subseteq I$, let $V_J = \langle e_j \mid j \in J \rangle$ (the linear span). $V_J$ is obviously $T$-invariant, as it has a basis of eigenvectors of $T$. By Zorn's Lemma, there exists a maximal $J$ such that $V_J \cap W = \{0\}$; we claim that $V = W \oplus V_J$.

Let $i \not \in J$; by maximality, $\left( V_J \oplus \langle e_i \rangle \right) \cap W \neq \{0\}$, so there exists $w \in W$ such that $w = z + ae_i$ for some $z \in V_J$, $a \neq 0$ in the base field (if $a=0$ then $V_J \cap W \neq \{0\}$, a contradiction). By rescaling $w$ and $z$ we can assume $a=1$. But now $e_i = w-z \in W \oplus V_J$ for every $i \in I$, so $V = W \oplus V_J$, which means that $V_J$ is an invariant complement for $W$.

We want to show that $\left( V_J \oplus \langle e_i \rangle \right) \cap W$ is a 1-dimensional invariant subspace of $W$. It is an invariant subspace because it is intersection of invariant subspaces. Suppose that there are two linearly independent vectors $w_1 = z_1 + a_1 e_i$, $w_2 = z_2 + a_2 e_i \in \left( V_J \oplus \langle e_i \rangle \right) \cap W$ (in the same notation as above). Then $a_1 w_2 - a_2 w_1 = a_1 z_2 - a_2 z_1 \in V_J \cap W$, so it must be $0$, a contradiction because we supposed they were linearly independent.

It follows that $\left( V_J \oplus \langle e_i \rangle \right) \cap W$ is spanned by some $w_i \in W$ that is an eigenvector for $T$; thus $\{ w_i \mid i \in I \setminus J \}$ is a basis of T-eigenvectors for $W$ and so the compression of $T$ to $W$ is diagonalizable.

Alessandro Iraci

1.6K

Matchmaticians Alumnus

0

Hey, thank you very much. Two questions, by "base field" you mean F? What do you mean by rescaling w and z? Just making sure I got everything.
Alessandro Iraci

0

By base field I mean the field V is a vector space on; you didn't give it a name, but I guess F should do. By "rescaling w and z" I mean you can divide both w and z by a and get two vectors w' and z' such that w' = z' + e_i, and then "rename" w' and z' as w and z again.
Alessandro Iraci

0

I hope it is clear, if not just feel free to ask again!
Matchmaticians Alumnus

0

Hey. A bit late, but I was revising this today and I have another question, if you have the time to break it down for me. Why exactly is $\{w_i : i\not\in J\}$ a basis for $W$?
Matchmaticians Alumnus

0

If you'd prefer, I could go to your profile and ask a question directly.
Matchmaticians

0

Outside user interactions are prohibited. Alessandro Iraci will be notified automatically when you post a question.
Matchmaticians Alumnus

0

Oh, sorry about that. I thought the ask a question button in a user's profile was asked exclusively to said user. Just finishing typing up the question. Would it be against the rules to post the answer Alessandro Iraci gave to ask for clarification on it?
Matchmaticians

0

No, it is fine.
Matchmaticians Alumnus

0

ok, thank you!
Alessandro Iraci

0

Hi, I'm in the EU time zone, so I was sleeping when you posted your comment. I would have been happy to clarify, but since you asked another question it's more convenient if I answer that directly (so I can type math).
Matchmaticians Alumnus

0

No worries, just pack 5 bucks worth of detail in there ;). I think I actually understood it a week ago, didn't even think to ask, but I'm drawing a blank revising now. Thank you!
Alessandro Iraci

0

I took it personally and wrote way more than what I think 5$ of details are worth, I hope it's fine now! :)

$[Linear Algebra] $T$-invariant subspace$

Answer

Related Questions

Search