The matrix trace is
Using matrix multiplication and interchanging dummy indices,
The proposed noncyclic interchange is false. With matrix units,
one has
Trace is invariant under cyclic permutations, but an arbitrary swap need not be cyclic.
The characteristic polynomial of is
A scalar is an eigenvalue when there is a nonzero vector such that
Such a is an eigenvector belonging to . Equivalently, is a root of and the corresponding eigenvectors are the nonzero elements of .
Consider the real polynomial
The hypothesis says , so is not identically zero. A nonzero polynomial has only finitely many roots, whereas is infinite. Hence there is a real with
Thus is invertible, as summarized by the real parameter avoiding a singular matrix pencil.
Write the complex similarity matrix as
with real. From we obtain , and equating real and imaginary parts gives
Therefore
for every real .
Now is a real polynomial and
Thus some real has . Taking , we have a real invertible matrix satisfying , or
This proves real similarity from complex similarity.
Eigenvectors belonging to distinct eigenvalues are linearly independent. Indeed, suppose
is a shortest nontrivial relation, where and the are distinct. Applying gives
a shorter nontrivial relation, a contradiction.
Thus distinct eigenvalues provide a basis of eigenvectors, proving that is diagonalizable over . If is real and all its eigenvalues are real, each real singular matrix has a nonzero real null vector. Choosing one for each eigenvalue gives a real eigenbasis. Hence the real matrix is diagonalizable over . This is the distinct eigenvalues imply diagonalizability criterion.

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