# Determine the center of $GL_n(\mathbb{R})$ [Artin 2.5.7] (similar to Axler 3.D.16)

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.

• 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. – Arturo Magidin Jul 22 at 1:54
• 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. – Arturo Magidin Jul 22 at 1:59
• 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 $T$. – Arturo Magidin Jul 22 at 2:03
• 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. – Arturo Magidin Jul 22 at 2:04
• 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. – Arturo Magidin Jul 22 at 2:26