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I encountered the following statement in the context of the Hecke algebra on the space of cusp forms, left without proof:

Let there be a family of commutative self-adjoint operators on a finite dimensional vector space $V$. Then there exists a basis of $V$ consisting of functions which are eigenfunctions for all the operators.

I would appreciate a proof of this, as elementary as possible, or a reference. I only know basic functional analysis and no spectral theory so I don't know where to start.

bsbb4
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1 Answers1

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Starting point: A normal matrix is diagonalizable (over the complex numbers).

Then any two commuting self-adjoint matrices are simultaneously diagonalisable (via a common set of eigenvectors).

Proof: Let $A,B$ be the two matrices; let $T:=A+iB$; this is a normal matrix ($T^*T=TT^*$) hence diagonalisable to $P^{-1}TP=C+iD$, hence $P^{-1}AP=C$, $P^{-1}BP=D$ (note that the real and imaginary parts are unique). Writing $AP=PC$ in terms of the column vectors, we find $Ap_i=c_ip_i$, and similarly, $Bp_i=d_ip_i$, so $p_i$ are a common set of eigenvectors for $A,B$.

Corollary: Any set of commuting self-adjoint matrices have a common set of eigenvectors.

Proof: They all commute with one of them $A$, hence have the same eigenvectors.

Chrystomath
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