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There have been some other, related questions: here, here, and here, but I wanted to verify whether I have any errors in my proof. While this question was taken from Artin's Algebra [2.5.7], it reminded me of a related question in Axler's Linear Algebra Done Right [3.D.16], so I tried to solve it using a linear algebra approach.

Exercise [Artin 2.5.7]: Determine the center of $GL_n(\mathbb{R})$. Hint: You are asked to determine the invertible matrices $A$ that commute with every invertible matrix $B$. Do not test with a general matrix $B$. Test with elementary matrices.

Original Proof (updated proof below): We prove the contrapositive: if $T \in \mathcal{L}(V)$ is not a scalar multiple of the identity then $ST \neq TS$ for every invertible $S \in \mathcal{L}(V)$. Since $T$ is not a scalar multiple of the identity, there exists some $v \in V$ for which $v, Tv$ are linearly independent. By 2.33 in Axler's LADR we can extend to a basis of $V: v, Tv, v_3, \ldots, v_n$. By 3.5 in LADR, we can define a $S \in \mathcal{L}(V)$ such that $Sv = v$, $S(Tv) = cTv$ where $c \neq 0$ and $c \neq 1$, and $Sv_j = v_j$ for $j=3,\ldots,n$. Then

$Tv \neq cTv \Rightarrow T(Sv) \neq S(Tv) \Rightarrow (TS)v \neq (ST)v$

Since the range of $S$ is n-dimensional, $S$ is surjective. By 3.69 in LADR, we know then that $S$ is also invertible. Therefore we've proven that if $ST = TS$ for every invertible map $S \in \mathcal{L}(V)$, then $T$ is a scalar multiple of the identity.

My Questions: Some things I'd like verification on

  • Did I properly define a $S \in \mathcal{L}(V)$ such that $S$ is invertible but is not commutative with $T$?
  • Is it correct to assume that "there exists some $v \in V$ for which $v, Tv$ is linearly independent" if $T$ is not a scalar multiple of $I$? I didn't include the proof but it would look something like:

Suppose for all $v \in V$, $Tv = {\alpha}v$ where $\alpha$ is some constant and $v_1, v_2, \ldots, v_n$ form a basis of $V$. Then $Tv = {\alpha}v = \alpha(c_1v_1 + c_2v_2 + \ldots + c_nv_n)$, but also $Tv = c_1Tv_1 + c_2Tv_2 + \ldots + c_nTv_n = c_1\beta_1v_1 + c_2\beta_2v_2 + \ldots + c_n\beta_nv_n$, which implies that $\alpha = \beta_1 = \beta_2 = \ldots = \beta_n$.


Updated Proof (based on feedback):

We prove that if $T \in \mathcal{L}(V)$ is not a scalar multiple of the identity then there exists an invertible $S \in \mathcal{L}(V)$ such that $ST \neq TS$.

Lemma. If $T$ is not a scalar multiple of the identity, there exists some $v \in V$ for which $v, Tv$ are linearly independent.

By 2.33 in Axler's LADR we can extend $v, Tv$ to a basis of $V: v, Tv, v_3, \ldots, v_n$. By 3.5 in LADR, we can define a $S \in \mathcal{L}(V)$ such that $Sv = v$, $S(Tv) = 2Tv$, and $Sv_j = v_j$ for $j=3,\ldots,n$. Then

$Tv \neq 2Tv \Rightarrow T(Sv) \neq S(Tv) \Rightarrow (TS)v \neq (ST)v$

Since the range of $S$ is n-dimensional, $S$ is surjective. By 3.69 in LADR, we know then that $S$ is also invertible. Therefore we've proven that if $ST = TS$ for every invertible map $S \in \mathcal{L}(V)$, then $T$ is a scalar multiple of the identity. It follows that every scalar multiple of the identity is central to $GL_n(\mathbb{R})$.

Proof of Lemma: Suppose for all $v \in V$, $Tv = a_vv$ for some scalar $a_v$ that depends on $v$. If not all $a_v$ are equal, then let $v_1, v_2 \in V$ such that $a_{v_1} \neq a_{v_2}$ (necessarily independent). It follows that $T(v_1 + v_2) = T(v_1) + T(v_2) = a_{v_1}v_1+a_{v_2}v_2$ and $T(v_1 + v_2) = a_{v_1+v_2}(v_1+v_2)$. But since $a_{v_1}v_1+a_{v_2}v_2 \neq a_{v_1+v_2}(v_1+v_2)$ (would imply $a_{v_1} = a_{v_1+v_2} = a_{v_2}$) we have a contradiction.

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    $\begingroup$ What you state is not the contrapositive, is not what you've proven, and is false as stated! The contrapositive of "if $T$ is central then it is a scalar multiple of the identity" is that if $T$ is not a scalar multiple of the identity, then there exists an invertible matrix $S$ such that $TS\neq ST$ (there always exist invertible matrices that do commute with $T$; for example, $T$ itself, and any polynomial in $T$, for starters). Also, this is not "the contrapositive" of the given statement, because you were told to determine/describe something, not to prove an implication. $\endgroup$ – Arturo Magidin Jul 22 at 1:54
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    $\begingroup$ You actually defined infinitely many $S$, not a single one. You could define a single one by picking a specific value of $c$, but what you do is fine. $\endgroup$ – Arturo Magidin Jul 22 at 1:59
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    $\begingroup$ Your proof of the proposition about $v$, $Tv$ is ambiguous: you don't specify if $a$ is the same for all $v$ or depends on $v$. Perhaps easier would be to say: if $v,Tv$ is linearly dependent for all $v$, then for all $v\neq 0$, $Tv=a_vv$ for some scalar $a_v$ that depends on $v$. If they are not all equal, pick $v_1,v_2$ (necessarily linearly independent) with $a_{v_1}\neq a_{v_2}$. Then $T(v_1+v_2) = a_1v_1+a_2v_2\neq k(v_1+v_2)$ for any $k$, contradicting our hypothesis about $T$. $\endgroup$ – Arturo Magidin Jul 22 at 2:03
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    $\begingroup$ Yes, "we prove that" would be better. That would establish that any central matrix is a scalar multiple of the identity. Then you would want to make the trivial observation that any scalar multiple of the identity is central. It's okay to define multiple $S$, but if you say you are going to define one, it is best if you actuallyt define one and not an entire set of such matrices. $\endgroup$ – Arturo Magidin Jul 22 at 2:04
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    $\begingroup$ Whether it is included or not depends on the intended audience. If I were grading this in a linear algebra course (or an introductory abstract algebra course), I would expect a proof. It can be done either as a "Claim" at the top, or if you wish to postpone the proof you should indicate, perhaps parenthetically, that a proof will be provided below (in which case, you can state it as a lemma). But you shouldn't leave the reader wondering if you are going to prove it or not. $\endgroup$ – Arturo Magidin Jul 22 at 2:26

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