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Solution to Linear Algebra Hoffman & Kunze Chapter 6.5


Exercise 6.5.1

Find an invertible real matrix P such that P1AP and P1BP are both diagonal, where A and B are the real matrices
(a)A=[1202],B=[3801](b)A=[1111],B=[1aa1].Solution: The proof of Theorem 8 shows that if a 2×2 matrix has two characteristic values then the P that diagonalizes A will necessarily also diagonalize any B that commutes with A.

(a) Characteristic polynomial equals (x1)(x2). So c1=1, c2=2.
c1:[0202][10]=[00]c2:[1200][21]=[00]So P=[1201] and P1=[1201].
P1AP=[1002],P1BP=[3001](b) Characteristic polynomial equals x(x2). So c1=0, c2=2. c1:[1111][11]=[00]c2:[1111][11]=[00]So P=[1111] and P1=[1/21/21/21/2].P1AP=[0002],P1BP=[1a001+a].


Exercise 6.5.2

Let F be a commuting family of 3×3 complex matrices. How many linearly independent matrices can F contain? What about the n×n 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 n24+1. A simpler proof was published in 1998 by M. Mirzakhani in the American Mathematical Monthly.


Exercise 6.5.3

Let T be a linear operator on an n-dimensional space, and suppose that T has n distinct characteristic values. Prove that any linear operator which commutes with T is a polynomial in T.

Solution: Since T has n distinct characteristic values, T is diagonalizable (exercise 6.2.7, page 190). Choose a basis B for which T is represented by a diagonal matrix A. Suppose the linear transformation S commutes with T. Let B be the matrix of S in the basis B. Then the ij-th entry of AB is aiibij and the ij-th entry of BA is ajjbij. Therefore if aiibij=ajjbij and aiiajj, then it must be that bij=0. So we have shown that B must also be diagonal. So we have to show there exists a polynomial such that f(aii)=bii for all i=1,,n. By Section 4.3 there exists a polynomial with this property.


Exercise 6.5.4

Let A, B, C, and D be n×n complex matrices which commute. Let E be the 2n×2n matrix
E=[ABCD].Prove that detE=det(ADBC).

Solution: If A is invertible, then[ABCD][InA1B0In]=[A0CDCA1B].Hence detE=detAdet(DCA1B)=det(ADACA1B).Since A and C commute, we have ADACA1B=ADCB. Thus detE=det(ADACA1B)=det(ADBC).If A is not invertible, then we set At=A+tIn, where t is a complex number. LetEt=[AtBCD].Clearly, for enough small t, At is invertible (detAt is a polynomial of t which can only have finitely many solutions). Moreover, At and C commute. Hence by the case we proved, we have detEt=det(AtDBC) for sufficiently small t. But both sides are polynomials in t, hence we can take t0 and the limit should be the same. That is detE=det(ADBC).


Exercise 6.5.5

Let F be a field, n a positive integer, and let V be the space of n×n matrices over F. If A is a fixed n×n matrix over F, let TA be the linear operator on V defined by TA(B)=ABBA. Consider the family of linear operators TA obtained by letting A vary over all diagonal matrices. Prove that the operators in that family are simultaneously diagonalizable.

Solution: If we stack the cloumns of an n×n matrix on top of each other with column one at the top, the matrix of TA in the standard basis is then given by
[AAA].Thus if A is diagonal then TA is diagonalizable.

Now TATB(C)=ABCACBBCA+CBA and TBTA(C)=BACBCAACB+CAB. Therefore we must show that BAC+CAB=ABC+CBA. The i,j-th entry of BAC+CAB is cij(aiibii+ajjbjj). And this is exactly the same as the i,j-th entry of ABC+CBA. Thus TA and TB commute. Thus by Theorem 8 the family can be simultaneously diagonalized.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
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