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[SCHUR's LEMMA] Let $A$ be an $n \times n$ matrix. If $A B=B A$ for all invertible matrices $B$, show that $A=c I$ for some scalar $c$.
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To prove this statement, we will use Schur's lemma as follows. Let $V$ be the space of all $n \times n$ matrices and let $G$ be the group of all invertible $n \times n$ matrices under matrix multiplication. Then $V$ is a representation of $G$, and the action of $G$ on $V$ is given by

$g \times A = g A g^{-1}$

for all $g$ in $G$ and $A$ in $V$.

Now, consider the linear transformation $T: V \rightarrow V$ defined by

$T(A) = AB$

for all $A$ in $V$. We will show that $T$ is a conjugacy between the representation $V$ and itself, so by Schur's lemma, $T$ must be unitarily equivalent to the identity transformation.

To show that $T$ is a conjugacy, we need to verify that

$T(g \times A) = g \times T(A)$

for all $g$ in $G$ and $A$ in $V$. Substituting the definitions of $g \times A$ and $T(A)$, we get

$T(g A g^{-1}) = g AB g^{-1}$

which is true because matrix multiplication is associative. Therefore, $T$ is a conjugacy between the representation $V$ and itself, and by Schur's lemma, $T$ must be unitarily equivalent to the identity transformation.

This means that there exists a unitary matrix $U$ such that

$AB = U AU^{-1}$

for all $A$ in $V$. Taking the trace of both sides of this equation, we get

$tr(AB) = tr(U AU^{-1})$

Since the trace is invariant under unitary transformations, it follows that

$tr(AB) = tr(AU^{-1}U) = tr(A)$

for all $A$ in $V$. In particular, this equation holds for $A = I$, the identity matrix. Therefore,

$tr(B) = tr(I) = n$

for all invertible $B$.

Now, consider the linear transformation $S: V \leftarrow V$ defined by

$S(A) = AB - BA$

for all $A$ in $V$. Then

$tr(S(A)) = tr(AB - BA)$

$= tr(AB) - tr(BA)$

$= n - n = 0$

for all $A$ in $V$. Therefore, $S$ is the zero transformation, which means that

$AB = BA$

for all $A$ in $V$. In particular, this equation holds for $A = B^{-1}$, so

$BB^{-1} = B^{-1}B$

which implies that $B^{-1}B = I$. Therefore, $B$ is unitary, which means that

$AB = B A$

for all unitary $B$.

Finally, consider the linear transformation $T': V \rightarrow V$ defined by

$T'(A) = AB - A$

for all $A$ in $V$. Then

$T'(AB - A)$

$= (AB - A)B - (AB - A)$

$= -AB + AB = 0$

for all unitary $B$, so $T'$ is the zero transformation. This means that

$AB = A$

for all unitary $B$, which implies that $A$ is a scalar multiple of the identity matrix. Therefore,

$A = c I$

for some scalar $c$. This completes the proof.
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