Let and let . Prove that every element of is conjugate to some element of . Deduce that is the union of conjugates of . [Hint: Show that every element of has an eigenvector.]

Let ; then .

- If , then .
- If , let . Note that , so that ; moreover, . Then .
- If and , let . Note that , so that . Evidently . Note that .
- Suppose now that and . Let , where . Note that , so that . Evidently . Moreover, we see that .

Thus every element of is conjugate to some element of ; clearly then .

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## Comments

Is there any method using eigenvector or diagonalization? I almost forgot linear algebra, so i don’t know if it is possible. If an invertible matrix has a distinct two eigenvalue, it is diagonalizable so it is a conjugate of a diagonal matrix, which is an element of H. But If an invertible matrix has a repeated eigenvalue, is there any way to proceed?

There is a much easier way- for one thing, it’s enough to note that every invertible matrix has an invertible Jordan canonical form. This is basically the method you mentioned in a more general context. With a repeated eigenvalue, a 2×2 matrix is ‘almost’ diagonalizable, in the sense that we can make the entry here equal to 1.

I try to make these solutions refer back only to things already proved in D&F, so I wrote the more explicit solution here.