Exercise 6.5.1
Find an invertible real matrix such that and are both diagonal, where and are the real matrices
Solution: The proof of Theorem 8 shows that if a matrix has two characteristic values then the that diagonalizes will necessarily also diagonalize any that commutes with .
(a) Characteristic polynomial equals . So , .
So and .
(b) Characteristic polynomial equals . So , . So and .
Exercise 6.5.2
Let be a commuting family of complex matrices. How many linearly independent matrices can contain? What about the case?
Solution: This turns out to be quite a hard question, so I’m not sure what Hoffman & Kunze had in mind. But there’s a general theorem from 1905 by I. Schur which says the answer is . A simpler proof was published in 1998 by M. Mirzakhani in the American Mathematical Monthly.
Exercise 6.5.3
Let be a linear operator on an -dimensional space, and suppose that has distinct characteristic values. Prove that any linear operator which commutes with is a polynomial in .
Solution: Since has distinct characteristic values, is diagonalizable (exercise 6.2.7, page 190). Choose a basis for which is represented by a diagonal matrix . Suppose the linear transformation commutes with . Let be the matrix of in the basis . Then the -th entry of is and the -th entry of is . Therefore if and , then it must be that . So we have shown that must also be diagonal. So we have to show there exists a polynomial such that for all . By Section 4.3 there exists a polynomial with this property.
Exercise 6.5.4
Let , , , and be complex matrices which commute. Let be the matrix
Prove that .
Solution: If is invertible, thenHence Since and commute, we have . Thus If is not invertible, then we set , where is a complex number. LetClearly, for enough small , is invertible ( is a polynomial of which can only have finitely many solutions). Moreover, and commute. Hence by the case we proved, we have for sufficiently small . But both sides are polynomials in , hence we can take and the limit should be the same. That is .
Exercise 6.5.5
Let be a field, a positive integer, and let be the space of matrices over . If is a fixed matrix over , let be the linear operator on defined by . Consider the family of linear operators obtained by letting vary over all diagonal matrices. Prove that the operators in that family are simultaneously diagonalizable.
Solution: If we stack the cloumns of an matrix on top of each other with column one at the top, the matrix of in the standard basis is then given by
Thus if is diagonal then is diagonalizable.
Now and . Therefore we must show that . The -th entry of is . And this is exactly the same as the -th entry of . Thus and commute. Thus by Theorem 8 the family can be simultaneously diagonalized.