Let Y be an arbitrary set of commuting matrices in \(M _ { n } ( \mathbf { C } )\) \(( \mathrm { i . e . , } A B = B A\) for all \(A , B \in Y )\) Prove that there exists a non-zero vector \(v \in \mathbf { C } ^ { n }\) which is a common eigenvector of all elements of Y.
Solution
We prove the assertion by induction on \(n\).
1. The assertion holds when \(n=1\).
Every linear operator on the one-dimensional space \(\CC\) acts by a scalar, so every nonzero vector is an eigenvector of every member of \(Y\). This also covers the case \(Y=\varnothing\).
2. Assume \(n>1\) and the assertion is known in every smaller positivedimension. If \(Y=\varnothing\), then every nonzero vector in \(\CC^n\) is acommon eigenvector.
There are no eigenvector conditions to check when the family is empty. Thus the assertion is vacuous in this case.
3. Suppose \(Y\neq\varnothing\). If every member of \(Y\) is a scalarmatrix, then every nonzero vector in \(\CC^n\) is a common eigenvector.
For each \(A\in Y\), write \[ A=\lambda_A I. \] Then every nonzero \(v\in\CC^n\) satisfies \[ Av=\lambda_Av. \]
4. Suppose some \(A\in Y\) is not scalar. Then \(A\) has an eigenspace\[ E_\lambda=\ker(A-\lambda I) \]satisfying\[ 0<\dim E_\lambda<n. \]
Because the ground field is \(\CC\), the characteristic polynomial of \(A\) has a root \(\lambda\), so \(E_\lambda\neq\{0\}\). If \(E_\lambda=\CC^n\), then \(A=\lambda I\), contrary to the choice of \(A\). Thus the eigenspace is proper.
5. The subspace \(E_\lambda\) from step 4 is invariant under every\(B\in Y\).
Let \(v\in E_\lambda\). Since \(A\) and \(B\) commute, \[ A(Bv) = B(Av) = B(\lambda v) = \lambda Bv. \] Hence \(Bv\in E_\lambda\).
6. The restricted family\[ \{B|_{E_\lambda}:B\in Y\} \]is a commuting family of endomorphisms of the smaller complex vectorspace \(E_\lambda\).
Step 5 makes every restriction well-defined. For \(B,C\in Y\), \[ (B|_{E_\lambda})(C|_{E_\lambda}) = (BC)|_{E_\lambda} = (CB)|_{E_\lambda} = (C|_{E_\lambda})(B|_{E_\lambda}), \] because the members of \(Y\) commute.
7. There exists a nonzero vector \(v\in E_\lambda\) that is aneigenvector of every \(B\in Y\).
6. It yields a nonzero \(v\in E_\lambda\) such that, for every\(B\in Y\),\[ Bv=\mu_Bv \]for some scalar \(\mu_B\in\CC\).
8. Therefore every commuting family \(Y\subseteq M_n(\CC)\) has acommon nonzero eigenvector.
Steps 2 and 3 settle the empty-family and scalar-only cases. In the remaining case, step 7 gives a vector that is an eigenvector of every member of \(Y\). Since \(v\in E_\lambda\), it is in particular an eigenvector of the chosen matrix \(A\) as well.
9. Q.E.D.
Step 8 completes the induction.