MASA's over finite dimensional Hilbert spaces contain non trivial projections

Question

I am dealing with an exercise in (elementary) operator theory and I am trying to prove that if $\mathcal{A}$ is a maximal abelian, self-adjoint subalgebra of $M_n(\mathbb{C})$ (where $n>1$) then:

$\mathcal{A}$ contains a non-zero projection, different than the unit $I_n$.

I know that vN algebras are rich in projections and since we are in finite dimension, $\mathcal{A}$ is a vN algebra. The thing is I am sure that I can prove this with elementary arguments, but I can't get around it. Could somebody help me out?

Answer 1

Fix an element $a\in\mathcal{A}$. Then $a$ is normal, so by the spectral theorem we can conjugate by a unitary and assume $a$ is a diagonal matrix. If the diagonal entries of $a$ are not all the same, we can find a polynomial $p$ which sends some of those diagonal entries to $1$ and the others to $0$, and then $p(a)\in\mathcal{A}$ will be a nontrivial projection.

The only case not covered by this argument is if all the diagonal entries of $a$ are the same, and this is true for every $a\in\mathcal{A}$. But then $\mathcal{A}$ is just the multiples of the identity, and fails to be maximal since it can be enlarged to all the diagonal matrices.

(Note that essentially the same argument works for masas in $B(H)$ for any Hilbert space $H$, not necessarily finite dimensional; just instead of a polynomial $p$ you need to use the Borel functional calculus to get a nontrivial spectral projection from $a$ unless $a$ is a multiple of the identity.)

Answer 2

Polynomial functional calculus should be enough. Let $\mathcal B$ be a (strict or not) subalgebra of $\mathcal A$, so that it contains a self-adjoint element $b=b^*$ with a non-trivial spectrum. We separate this set of eigenvalues of $b$ in two parts, and find a polynomial $P$ with real coefficients which is $0$ on the one part and $1$ on the other part. Because of $P^2=P$ on the spectrum of $b$, we have for $p=P(b)$ the relation $$ p^2 = (P(b))^2=(P^2)(b)=P(b)=p\ . $$ Because $P$ has real coefficients, $$ p^*=P(b)^*=\bar P(b)=P(b)=b\ . $$ (Well, $p$ is the projection on the direct sum of the eigenspaces of $b$ corresponding to the $1$-part.)

It remains to show that a subalgebra $\mathcal B$, which has the maximality property from the OP admits such a $b$. Suppose this is not the case, than for all $x\in \mathcal B$ we have that $x+x^*$ is a multiple of the identity. Let $\tau$ be the normed trace, $\tau(1)=1$. We get $x+x^*=2\tau(x)$. So $\mathcal B$ should be a subalgebra of a very special vectorial subspace $V$ of $\mathcal B$, with an "anti-mirroring" of values outside the diagonal. However, taking the image of $x\to xx^*$ on $V$ brings us outside $V$ for an $n\ge 2$. Contradiction.