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An eigenvalue of $C$ is a scalar $\lambda$ for which a nonzero vector $v$ satisfies $Cv=\lambda v$. Equivalently, $\det (\lambda I-C)=0$, because the homogeneous linear system has a nontrivial solution exactly when its coefficient matrix is singular. This characteristic polynomial is the monic quadraticA nonzero quadratic has at most two distinct roots, proving the eigenvalue bound. Eigenvalues and diagonalization are taken over $\mathbb{C}$, also for a matrix with real entries.

For the matrix commutator, the cyclic trace identity follows directly from its entries:Hence $\mathrm{tr}[A,B]=0$. Factoring the characteristic polynomial as $(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)$ and comparing the coefficient of $\lambda$ gives $\mathrm{tr}C={\lambda }_1+{\lambda }_2$, with the eigenvalues counted with algebraic multiplicity. If both eigenvalues are zero, trace and determinant are zero. The Cayley-Hamilton theorem then gives $C^2=0$; no diagonalizability assumption is needed.

Now put $M=[A,B]$. Since its trace is zero, its characteristic polynomial is ${\lambda }^2+\det M$ and the scalar square of a traceless two-by-two matrix identity givesThis proves the scalar-square assertion for both real and complex entries. If $\det M=0$, it gives $M^2=0$. If $\det M\ne 0$, the two eigenvalues are the distinct nonzero numbers $\pm \sqrt{-\det M}$. Their eigenvectors are linearly independent: a nonzero vector cannot satisfy two different eigenvalue equations. They form a basis, making $M$ a diagonalizable matrix. Thus either the commutator is diagonalizable over $\mathbb{C}$ or its square is zero. These alternatives need not be disjoint, since the zero matrix has both properties. Over $\mathbb{R}$ the diagonalization assertion would require real eigenvalues; complex diagonalization is essential to the unrestricted statement.