MA 52 May 13, 2006Chapters 5,6,7 ReviewSOLUTIONS PROBLEMS 1-20Math 52 Spring 20061. (a) Express the matrix A =!0.5 02 1.5"as a product SDS−1, whereD is a diagonal matrix.(b) Find a formula for Ak!10".(a) By inspe ction, the eigenvalues are λ1= 0.5 and λ2= 1.5 (since Ais lower triangular). Therefore D should beD =!0.5 00 1.5"However, to find S requires more work. The eigenspaces areEλ1= ker!0 02 1"= span#!1−2"$Eλ1= ker!−1 02 0"= span#!01"$Therefore an eigenbasis isB : "v1=!1−2", "v2=!01"andSB→std=!1 0−2 1"Sstd→B=!1 02 1"1MA 52 May 13, 2006Note that a different choice of basis for the e igenspaces will give adifferent (but still correct) set of change of basis matrices. That’sfine!We obtainA = SB→stdDSstd→B=!1 0−2 1"!0.5 00 1.5"!1 02 1"which multiplies out correctly.(b) The fast way to see this is to note that!01"is the eigenvectorfor λ2= 1.5. ThereforeAk!01"= λk2!01"=!01.5k"Alternatively, a more general approach is as follows. We have arepresentation A = SDS−1, so we see thatAk!01"= (SDS−1)k!01"= (SDS−1)(SDS−1) . . . (SDS−1)!01"= SDkS−1!01"=!1 0−2 1"!0.5k00 1.5k"!1 02 1"!01"=!01.5k"2. Compute the determinant of the following matrix:1 −1 −2 63 1 2 42 0 5 1−2 3 2 3.The determinant is 306. Try using row operations if you find theLaplace expansion tedious.2MA 52 May 13, 20063. Prove or disprove and salvage if possible:(a) Let A =!a bc d"and define the transpose of A by AT=!a cb d".Then A and AThave the same eigenvalues.(b) Every 3 × 3 matrix has at least one real eigenvalue.(c) A real number λ is an eigenvalue of A if and only if λ is an eigen-value of Anfor all positive integers n.(a) Proof: We have(A − λI)T=!a − λ cb d − λ"= AT− λIThereforedet((A − λI)) = det((A − λI)T) = det(AT− λI)and so the characteristic polynomials satisfy fA(λ) = fAT(λ) andhence both A and AThave the same eigenvalues.(b) Proof: The characteristic polynomial of a 3 × 3 matrix is a degreethree polynomial. Therefore it has at least one real root (by cal-culus – e.g. look at limit as λ → ∞ and λ → −∞). And hencethe matrix has at least one real eigenvalue.(c) Not true! Counterexample:Consider the scalingA =!2 00 2"This has eigenvalue λ = 2. ButA2=!4 00 4"has eigenvalue λ = 4. Therefore the “only if” part is false.3MA 52 May 13, 2006Salvage (corrected statement): A real number λ is an eigenvalue ofA if and only if λnis an eigenvalue of Anfor all p ositive integers n.Proof of Salvage: If λnis an eigenvalue of Anfor all positive in-tegers n, then in particular, λ1= λ is an eigenvalue of A1= A.This proves the “if” part.For the “only if”, assume that λ is an eigenvalue of A. Then bydefinition there is some vector "v such that A"v = λ"v. We proceedby induction on the claim that “Ak"v = λk"v”.Base case: The case n = 1 is exactly the statement of our assump-tion.Inductive step: Suppose we have proven the claim for n = k. ThenAk+1"v = AkA"v= Akλ"v by the initial assumption= λAk"v= λλk"v by the inductive hypothesis= λk+1"vAnd we have completed the proof.4. Either give an example exhibiting the stated properties or prove thatno such example exists.(a) Square matrices A and B with the same characteristic polynomialso that A is not similar to B.(b) A square matrix A which is not diagonalizable.(a) LetA =!1 00 1"B =!1 10 1"These two matrices have the same characteristic polynomial, i.e.(1 − λ)2. They both have eigenvalue λ = 1 with algebraic multi-plicity 2. However, λ has geometric multiplicity 2 for A and only4MA 52 May 13, 20061 for B. To see this geometrically, note that A is the identity,so the whole space is the eigenspace for λ = 1. However, B is ashear and shears are not diagonalizable. (If in doubt, verify thegeometric multiplicity by a calculation.)Note that if A and B are diagonalizable and have the same char-acteristic polynomial, then they have the same eigenvalues andare similar to the same diagonal matrix D. Therefore they aresimilar to each other (by transitivity of the equivalence relation“similar”). So to look for an example, we need matrices which arenot both diagonalizable.(b) The matrix B above is a good example of a nondiagonalizablesquare matrix.5. Assume thatA =3 4 3−1 −4 −51 8 9has characteristic polynomial 16 − 20t + 8t2− t3= −(t − 2)2(t − 4).Find the eigenvalues and eigenspaces of A.The eigenvalues are λ = 2 and λ = 4 of algebraic multiplicity 2 and 1respectively.The eigenspace E2can have geometric multiplicity 1 ≤ g ≤ 2.E2= ker1 4 3−1 −6 −51 8 7= span−11−1Note that this is clearly the kernel since the rank of the matrix is 1 (thefirst two columns are obviously independent since one is not a multipleof the other). So the geometric multiplicity of E2is g = 1.The eigenspace E4must have geometric multiplicity 1.5MA 52 May 13, 2006E4= ker−1 4 3−1 −8 −51 8 5= span1−23This matrix is not diagonalizable since the geometric multiplicities onlyadd up to 2.6. Let T : P2→ P2be defined by T (f) = f + f#+ f##. Find an eigenbasisfor T .First we must find the matrix for the transformation. We will use thebasisB : 1, t, t2The transformation acts as follows:T (a + bt + ct2) = a + bt + ct2+ b + 2ct + 2c = (a + b + 2c) + (b + 2c )t +ct2Therefore its matrix is[T ]B=1 1 20 1 20 0 1There is only one eigenvalue: λ = 1. The associated eigenspace isE1= ker0 1 20 0 20 0 0= span100This transformation does not have an eigenbasis, since there are notenough eigenvectors to form one.6MA 52 May 13, 20067. Let"v1=−122, "v2=1−10, "v3=010.These vectors form a basis of R3. (Note: you do not have to show this.)(a) Use the Gram-Schmidt process on these vectors to produc e anorthonormal basis of R3.(b) Let T : R3→ R3be the orthogonal projection of R3onto the sub-space spanned by "v1and "v2. Write down a matrix representing T.Hint: your work in part (a) might be useful.(a) The Gram-Schmidt process gives the result"u1=13−122, "u2=132−12, "u3=1322−1Remember to check the result by quickly dotting the vectors pair-wise in your head to make sure you get 0 or 1 where appropriate.(b) Recall that the Gram-Schmidt orthogonalization process tells usthatspan {"v1, "v2} = span {"u1, "u2}LetQ =13−1 22 −12