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Show that the only matrices that commute with all other matrices in $GL_{n}(\mathbb{R})$ will be multiples of the identity matrix. Or:

$Z(GL_{n}(\mathbb{R}))=\{\lambda I_{n}:\lambda \in \mathbb{R}\}$

One direction is clearly obvious. However, to show that every element $g$ such that $gh = hg$ for $h \in GL_{n}(\mathbb{R})$ has to be a multiple of the identity is causing me trouble as I don't know where to begin with. What would be a way to handle this problem? I was thinking of using contradiction, but that doesn't take me anywhere.

Thanks for the help.

user26857
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Guy
  • 989

2 Answers2

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The interesting (characterizing) property of scalar matrices (operators) is that every one dimensional subspace is invariant under them (i.e. eigenspace). This is easy to prove.

On the contrary, suppose that $A$ is an invertible matrix with some non-invariant one dimensional subspace, say $A(v_1)=v_2$ with $v_2\notin \langle v_1\rangle$.

Now the subspace $\langle v_1\rangle$ and $\langle v_2\rangle$ are different, and $A$ carries former to later. In particular, $\{v_1,v_2\}$ are linearly independent, hence $\{v_1+v_2, v_2\}$ are also linearly independent.

Let $B$ be an invertible matrix (transformation) such that $B(v_1)=v_1+v_2$ and $B(v_2)=v_2$. Then $$AB(v_1)=A(v_1+v_2)=v_2+A(v_2)$$ and $$BA(v_1)=B(v_2)=v_2.$$ So $AB=BA$ would imply that they agree at $v_1$, and hence $A(v_2)=0$, a contradiction (why?)

Hence, scalar matrices are in center, and non-scalar matrices are not in center of $GL_n$.

Groups
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Let $A\in Z(GL_n(\Bbb R))$. For fixed $i, j\in \{1,2,\ldots, n\}$ with $i\neq j$, let $X_{ij}\in GL_n(\Bbb R)$ be the matrix obtained from the identity matrix $I_n$ by interchanging the $i$th and $j$th rows of $I_n$. Then $AX_{ij}$ is the matrix obtained by interchanging the $i$th and $j$th columns of $A$, whereas $X_{ij}A$ is the matrix obtained by interchanging the $i$th and $j$th rows of $A$. The $(i,i)$-entry of $AX_{ij}$ is $a_{ij}$ and the $(i,i)$-entry of $X_{ij}A$ is $a_{ji}$. So since $AX_{ij} = X_{ij}A$, then $a_{ij} = a_{ji}$.

Now let $Y_{ij}\in GL_n(\Bbb R)$ be the matrix obtained from $X_{ij}$ be multiplying the $i$th row of $X_{ij}$ by $-1$. The $(i,i)$-entry of $AY_{ij}$ is $-a_{ij}$ and the $(i,i)$-entry of $Y_{ij}A$ is $a_{ji}$, so since $AY_{ij} = Y_{ij}A$, then $-a_{ij} = a_{ji}$.

Since $a_{ij} = a_{ji} = -a_{ij}$, then $a_{ij} = 0$. Furthermore, since the $(i,j)$-entry of $AX_{ij}$ is $a_{ii}$ and the $(i,j)$-entry of $X_{ij}A$ is $a_{jj}$, $a_{ii} = a_{jj}$. This shows that $A$ is a multiple of $I_n$.

kobe
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