Compute the Jordan canonical form over of the matrix whose every entry is equal to 1.

Let be an matrix. Note the following.

So we have , and thus the minimal polynomial of divides . Since and , the minimal polynomial of is .

Next we prove a lemma.

Lemma: Let be an matrix over a field containing the eigenvalues of , and let be an eigenvalue of with eigenspace . If the minimal polynomial of over is a product of distinct linear factors, then the dimension of (over ) is precisely the multiplicity of among the roots of the characteristic polynomial of . Proof: Let . As an module via , we have , where are the eigenvalues of . The eigenspace of consists precisely of those direct factors with , each of which has dimension 1 over .

For our given matrix , we claim that the eigenspace of the eigenvalue has dimension 1 over . To see this, let and suppose . Evidently, every entry of is , and so we have for all . Since , we have for all , and so . So has a basis containing one element, and so has dimension 1 over as desired.

By the lemma, has multiplicity 1 among the roots of the characteristic polynomial of . Since the remaining roots are 0, the invariant factors of are times and . The corresponding Jordan canonical form matrix is , with and otherwise.

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