# Schur's lemma

• Dec 15th 2011, 02:52 PM
ymar
Schur's lemma
I have begun reading Herstein's book, "Noncommutative Rings". I like it very much in general, great style. So if someone reading this hasn't seen it yet, I think I can recommend it.

I'm writing this because I've having difficulty understanding a part of the text about Schur's lemma. I understand the statement and the proof since they're straightforward. What I'm having trouble with is the special case he's talking about later on. He says this:

Quote:

Let $\displaystyle F$ be a field and let $\displaystyle F_n$ be the ring of all $\displaystyle n\times n$ matrices over $\displaystyle F.$ We consider $\displaystyle F_n$ as the ring of all linear transformations on the vector space $\displaystyle V$ of $\displaystyle n-$tuples of elements of $\displaystyle F.$ If $\displaystyle A$ is a subset of $\displaystyle F_n$ let $\displaystyle \overline{A}$ be the subalgebra generated by $\displaystyle A$ over $\displaystyle F.$ Clearly $\displaystyle V$ is a faithful $\displaystyle F_n,$ and so, $\displaystyle \overline{A}-$module. $\displaystyle V$ is, in addition, both a unitary and irreducible $\displaystyle F_n-$module.

We say the set of matrices $\displaystyle A$ is irreducible if $\displaystyle V$ is an irreducible $\displaystyle \overline{A}-$module. In matrix terms this merely says that there is no invertible matrix $\displaystyle S$ in $\displaystyle F_n$ so that

$\displaystyle S^{-1}aS=\left(\begin{array}{c|c} a_1 & 0 \\ \hline * & a_2 \end{array}\right)$

for all $\displaystyle a\in A.$

I understand everything but the last sentence. Are the blocks supposed to be square matrices? Of the same size? What is the quantificator for $\displaystyle a_1$ and $\displaystyle a_2?$
• Dec 15th 2011, 03:06 PM
NonCommAlg
Re: Schur's lemma
Quote:

Originally Posted by ymar
I have begun reading Herstein's book, "Noncommutative Rings". I like it very much in general, great style. So if someone reading this hasn't seen it yet, I think I can recommend it.

I'm writing this because I've having difficulty understanding a part of the text about Schur's lemma. I understand the statement and the proof since they're straightforward. What I'm having trouble with is the special case he's talking about later on. He says this:

I understand everything but the last sentence. Are the blocks supposed to be square matrices? Of the same size? What is the quantificator for $\displaystyle a_1$ and $\displaystyle a_2?$

strange, shouldn't the $\displaystyle *$ be $\displaystyle 0$?
• Dec 15th 2011, 03:08 PM
ymar
Re: Schur's lemma
PS: by blocks I meant $\displaystyle a_1$ and $\displaystyle a_2.$
• Dec 15th 2011, 03:15 PM
NonCommAlg
Re: Schur's lemma
Quote:

Originally Posted by ymar
PS: by blocks I meant $\displaystyle a_1$ and $\displaystyle a_2.$

• Dec 15th 2011, 03:20 PM
ymar
Re: Schur's lemma
Quote:

Originally Posted by NonCommAlg

Sorry, when I started writing, your question wasn't there and I missed it. It's $\displaystyle *$ in the book.
• Dec 15th 2011, 03:34 PM
NonCommAlg
Re: Schur's lemma
Quote:

Originally Posted by ymar
Sorry, when I started writing, your question wasn't there and I missed it. It's $\displaystyle *$ in the book.

$\displaystyle V$ being a reducible $\displaystyle \overline{A}$-module means $\displaystyle V$ has a non-trivial $\displaystyle \overline{A}$-submodule $\displaystyle V_1.$ obviously $\displaystyle V_1$ is also an $\displaystyle F$-subspace of $\displaystyle V$ and thus there exists an $\displaystyle F$-subspace $\displaystyle V_2$ such that $\displaystyle V = V_1 \oplus V_2.$ note that both $\displaystyle V_1,V_2$ are non-zero. now suppose that $\displaystyle B_1, B_2$ are some $\displaystyle F$-basis for $\displaystyle V_1, V_2$ respectively. then $\displaystyle B=B_1 \cup B_2$ is an $\displaystyle F$-basis for $\displaystyle V.$ clearly $\displaystyle ab_1 \in V_1$ for all $\displaystyle a \in A$ and $\displaystyle b_1 \in B_1$ because $\displaystyle V_1$ is an $\displaystyle \overline{A}$-module. if we look at the elements of $\displaystyle A$ as linear transformations, how will the matrix of an element $\displaystyle a \in A,$ with respect to $\displaystyle B$, look like?
• Dec 16th 2011, 06:11 AM
ymar
Re: Schur's lemma
Quote:

Originally Posted by NonCommAlg
$\displaystyle V$ being a reducible $\displaystyle \overline{A}$-module means $\displaystyle V = V_1 \oplus V_2$ for some non-zero $\displaystyle \overline{A}$-modules $\displaystyle V_1, V_2.$

Why is it so? Herstein's definition of irreducible module is this:

$\displaystyle M$ is said to be an irreducible $\displaystyle R-$module if $\displaystyle MR\neq (0)$ and if the only submodules of $\displaystyle M$ are $\displaystyle (0)$ and $\displaystyle M.$

When you say "reducible", do you mean not irreducible? Do you use the same definition of "irreducible" as Herstein does? I'm asking because I don't see why what you said is equivalent to $\displaystyle V$ either having a trivial scalar multiplication or having a nontrivial proper submodule.

Of course, if $\displaystyle V=V_1\oplus V_2,$ then $\displaystyle V_1$ is a nontrivial proper submodule. But I don't understand why the other implication would hold. Say $\displaystyle V$ contains a nontrivial proper submodule $\displaystyle V_1.$ Then it's not true for general modules that $\displaystyle V\cong V/V_1\oplus V_1,$ right?
• Dec 16th 2011, 01:29 PM
NonCommAlg
Re: Schur's lemma
Quote:

Originally Posted by ymar
Why is it so? Herstein's definition of irreducible module is this:

$\displaystyle M$ is said to be an irreducible $\displaystyle R-$module if $\displaystyle MR\neq (0)$ and if the only submodules of $\displaystyle M$ are $\displaystyle (0)$ and $\displaystyle M.$

When you say "reducible", do you mean not irreducible? Do you use the same definition of "irreducible" as Herstein does? I'm asking because I don't see why what you said is equivalent to $\displaystyle V$ either having a trivial scalar multiplication or having a nontrivial proper submodule.

Of course, if $\displaystyle V=V_1\oplus V_2,$ then $\displaystyle V_1$ is a nontrivial proper submodule. But I don't understand why the other implication would hold. Say $\displaystyle V$ contains a nontrivial proper submodule $\displaystyle V_1.$ Then it's not true for general modules that $\displaystyle V\cong V/V_1\oplus V_1,$ right?

ok, you're right. i made a small mistake! i was thinking of "completely reducible" not just "reducible"! it now makes sense why Herstein used $\displaystyle *$ instead of $\displaystyle 0$ at the bottom left of the matrix. anyway, i fixed the mistake and you should see my previous comment again!
• Dec 23rd 2011, 05:56 AM
ymar
Re: Schur's lemma
Quote:

Originally Posted by NonCommAlg
$\displaystyle V$ being a reducible $\displaystyle \overline{A}$-module means $\displaystyle V$ has a non-trivial $\displaystyle \overline{A}$-submodule $\displaystyle V_1.$ obviously $\displaystyle V_1$ is also an $\displaystyle F$-subspace of $\displaystyle V$ and thus there exists an $\displaystyle F$-subspace $\displaystyle V_2$ such that $\displaystyle V = V_1 \oplus V_2.$ note that both $\displaystyle V_1,V_2$ are non-zero. now suppose that $\displaystyle B_1, B_2$ are some $\displaystyle F$-basis for $\displaystyle V_1, V_2$ respectively. then $\displaystyle B=B_1 \cup B_2$ is an $\displaystyle F$-basis for $\displaystyle V.$ clearly $\displaystyle ab_1 \in V_1$ for all $\displaystyle a \in A$ and $\displaystyle b_1 \in B_1$ because $\displaystyle V_1$ is an $\displaystyle \overline{A}$-module. if we look at the elements of $\displaystyle A$ as linear transformations, how will the matrix of an element $\displaystyle a \in A,$ with respect to $\displaystyle B$, look like?

I understand now, except for one little doubt. Does the algebra $\displaystyle \overline{A}$ have to have identity? Herstein's rings in general needn't have it, and I don't know about his algebras. He sometimes assumes that the reader knows what he means by things and I think this is one of these cases, unless I missed something in the text. If $\displaystyle \overline{A}$ has identity, then $\displaystyle V_1$ is an $\displaystyle F-$subspace of $\displaystyle V,$ because for $\displaystyle f\in F$ and $\displaystyle v\in V_1,$ we have

$\displaystyle fv=\begin{bmatrix} f & & & \\ & f & & \text{\huge{0}} \\ \text{\huge{0}} & & \ddots & \\ & & & f \end{bmatrix}v \in V_1,$

because $\displaystyle V_1$ is an $\displaystyle \overline{A}-$module.
• Dec 24th 2011, 06:47 AM
NonCommAlg
Re: Schur's lemma
Quote:

Originally Posted by ymar
I understand now, except for one little doubt. Does the algebra $\displaystyle \overline{A}$ have to have identity?

yes because$\displaystyle \overline{A}$ is the algebra generated by $\displaystyle F$ and $\displaystyle A$ and so $\displaystyle \overline{A}$ contains a copy of $\displaystyle F$, i.e. $\displaystyle \overline{F} = \{aI : \ a \in F \},$ where $\displaystyle I$ is the identity matrix. thus $\displaystyle I$ is the identity element of $\displaystyle \overline{A}.$