Hill chapter 5.2: v and lambda are associated eigenvectors, eigenvalues of an nxn matrix A. In problems 23--28 you are asked to show directly that v is also an eigenvector of some related matrix B, and to find the eigenvalues. ***Note*** "Show directly" means you should evaluate Bv (where B is the matrix they are talking about in the problem) using Av=lambda*v. It really is that simple in all of these questions except for 25: **** If you see a v next to an A, replace it with v*lambda. **** 23. B=A^2 A^2v = AAv = Av*lambda = v*lambda*lambda So it works and the eigenvalue is lambda^2. 24. B=A^k, k a positive integer A..AAAv = A..AAv*lambda = A..Av*lambda*lambda = ....... = v*lambda^k. A more precise answer would use mathematical induction: Assuming A^(k-1) v = lambda^(k-1)*v, A^k v = A A^(k-1) v = A v*lambda^(k-1) = v*lambda*lambda^(k-1) = v*lambda^k.) Either way, conclude it works and the eigenvalue is lambda^k. 25. B=A^-1 A^-1 v = ? Well, there's no "Av" on the left here; need a new tactic. Take the equation Av=v*lambda and multiply on the left by A^-1: A^-1 A v = A^-1 v * lambda => v = A^-1 v * lambda => A^-1 v = v/lambda = v*(lambda^-1) So it works and the eigenvalue is lambda^-1. 26. B=A^k, k any integer If k>0, we did it. If k=0, A^k=I, and Av=1*v no matter what. If k<0, A^k=(A^-1)^(-k), so answers to 24 and 26 combine. Result: A^k v = lambda^k v: it's still true! 27. B=A-7I (A-7I)v = Av-7v = lambda*v-7v=(lambda-7)v So it works, and the eigenvalue is lambda-7. 28. B=A-aI (A-aI)v = Av-av = lambda*v-av=(lambda-a)v So it works, and the eigenvalue is lambda-a. 29. A square matrix B is called nilpotent if B^k=0 for some integer k>1. Show that 0 is the only eigenvalue of a nilpotent matrix. All eigenvalues are 0: Because of problem 24, if lambda is an eigenvalue of B then lambda^k is an eigenvalue of B^k=0. => lambda^k=0 => lambda=0 0 is an eigenvalue: [There are a million different ways to show this is true.] Proof 1: The characteristic equation is a polynomial of degree n. => It has some root(s); none are nonzero by above. => So the roots are zero! Proof 2: Same reasoning as above shows det(B)^k=det(B^k)=0 => det(B)=0 => B is singular => 0 is an eigenvalue Proof 3: Construct an eigenvector. Let v be a nonzero vector. Consider v, Bv, B^2v, ..., B^kv=0. At some point the vectors in this sequence change from nonzero to zero. At that point, B^j v != 0 and B(B^j v)=0 => B^j is an eigenvector with eigenvalue 0. 30. A square matrix C is called idempotent if C^2=C. What are the possible eigenvalues of an idempotent matrix? If Cv = lambda v, then by #24 C^2 v = lambda^2 v. But since C^2=C, C^2 v = C v = lambda*v. This can only happen if lambda^2=lambda => lambda is either 0 or 1. 31. Suppose that A is a 3x3 matrix with eigenvalues 0, 2, 4 and corresponding eigenvectors u1, u2, u3. The collection {u1, u2, u3} is necessarily linearly independent. So it forms a basis for R^3. Write x = a*u1 + b*u2 + c*u3. Then Ax = 0a*u1 + 2b*u2 + 4c*u3. (a) Find bases for NS(A) and CS(A). [HINT y in CS(A) => y=Ax] NS(A): 0a u1 + 2b u2 + 4c u3 = 0 iff b=c=0 iff (a,b,c)=(t,0,0). So NS(A) is spanned by u1! CS(A): y is in CS(A) <=> y = 2b u2 + 4b u3 for some b,c. So CS(A) = Span{u2, u3}. (b) Solve Ax=u2+u3. <=> 0a=0, 2b=1, 4c=4. Solution: x=(t)u1 + (1/2)u2 + (1/4)u3. (c) Show that Ax=u1 has no solution. Ax=u1 <=> 0a=1, 2b=0, 4c=0. That's impossible! 33 Note I = I^T So A^T - lambda*I = (A-lambda*I)^T So det(A^T - lambda*I) = det((A-lambda*I)^T) = det(A-lambda*I) => A and A^T have identical characteristic equations 35 (a) Write down two square matrices (not 1x1) at random. "With probability 1" you will find a counterexample; you just need to take the eigenvalues. (This is true in the same sense that if you put on a blindfold and randomly pick a point in the x-y plane, "with probability 1" you will NOT pick a point on the graph y=x^2.) (b) Ditto. (c) Depends on your example. (d) The sum of the eigenvalues of a matrix is its TRACE. The product of the eigenvalues of a matrix is its DETERMINANT. So these are just the identities Tr(A+B)=Tr(A)+Tr(B) det(AB)=det(A)*det(B)