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Proofs Homework Set 10

MATH 217 WINTER 2011

Due March 23

PROBLEM 10.1. Suppose thatA andB aren n matrices that commute (that is, AB = BA)and suppose thatB hasndistinct eigenvalues.

(a) Show that ifBv= vthenBAv= Av.

Proof. This follows from the fact that AB = BA. Indeed,

BAv= ABv= A(v) =Av

since scalar multiplication commutes with matrix multiplication.

(b) Show that every eigenvector forB is also an eigenvector forA.

Proof. Suppose v is an eigenvector ofBwith eigenvalue . By part (a), we have BAv= Av.So eitherAv= 0orAvis also an eigenvector ofB with eigenvalue. SinceB hasndistincteigenvalues, they all have multiplicity 1 which means that all of the eigenspaces ofB are

one-dimensional (see Theorem 7(b) in Section 5.3). Since v and Av both lie in the one-dimensional eigenspace ofB corresponding to the eigenvalue , v and Av must be linearlydependent. Since v= 0, this means that Av = v for some scalar . Therefore, v is aneigenvector ofAcorresponding to the eigenvalue .

(c) Show that the matrixAis diagonalizable.

Proof. SinceB has n-distinct eigenvalues, we know thatB is diagonalizable by Theorem 6 ofSection 5.3. ThereforeB has nlinearly independent eigenvectors v1, . . . ,vnby Theorem 5 ofSection 5.3. By part (b), the vectors v1, . . . ,vn are also eigenvectors ofA. Therefore,A hasn linearly independent eigenvectors, which means that A is diagonalizable by Theorem 5 ofSection 5.3.

(d) Show that the matrixAB is diagonalizable.

Proof. By the solution of part (c) and Theorem 5 of Section 5.3, we have

A= P DP1, B= P EP1


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whereD and Eare diagonal matrices and

P =v1 vn

.

Note that we get the same matrix P forA and B since v1, . . . ,vn are eigenvectors of bothAandB . However, the eigenvalues corresponding to these eigenvectors may be different for A

andB so we get different diagonal matricesD and E.From this, we see that

AB= P DP1P EP1 =PDEP1,

which shows thatAB is diagonalizable sinceDEis a diagonal matrix.

PROBLEM10.2. The sequence of Lucas numbersis defined by the recurrence formula

L0= 2, L1= 1, Ln= Ln1+ Ln2for n 2.Thus, the sequence starts like this

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, . . .

In this problem, you will use linear algebra to find an explicit formula for Ln, the nth Lucasnumber.

LetA=

1 11 0

.

(a) Show thatA

LnLn1

=

Ln+1

Ln

for eachn 1.

Proof. We have

A Ln

Ln

1=

1 1

1 0

Ln

Ln

1=

Ln+ Ln1

Ln =

Ln+1

Ln

sinceLn+1= Ln+ Ln1by definition of the Lucas numbers.

(b) Use part (a) to prove by induction thatAn

12

=

Ln+1

Ln

holds for everyn 0.

Proof. Base case. First note thatA0 =Iby definition. Therefore, A0

12

=

12

=

L1L0

by

definition ofL0and L1.

Induction step. Suppose we know thatAn

12

=

Ln+1

Ln

; we want to show that An+1

12

=

Ln+2Ln+1

. Well, sinceAn+1 =AAn, we see that

An+1

12

= A

An

12

= A

Ln+1

Ln

by the Induction Hypothesis. But A

Ln+1

Ln

=

Ln+2Ln+1

by part (a), and so An+1

12

=

Ln+2Ln+1

as claimed.


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(c) Find the two eigenvalues for A. Call the positive one (this is the Greek letter phi) andverify that the negative one is equal to 1/.

Proof. The characteristic equation ofAis

det(A I) = det 1 11 = (1 )() 1 =2 1.The quadratic formula gives the two roots

1 +

5

2

152

.

The first root is clearly positive, and so = 12

(1 +

5) = 1.61803 . . . (This is the famous

Golden Ratio!) The second root is negative, 12

(15) = 0.61803 . . .The fact that the second root is

1/can be checked by multiplying the two numbers:

1 +

5

2

1 52

=1 5 + 5 5

4 =

44

= 1.

(d) Find eigenvectors corresponding to these two eigenvaluesand 1/.

Proof. For the first eigenvalue, we have

A I=

1 11

1 0 0

and thus we find the corresponding eigenvector

1

.

For the second eigenvalue 1/, we have

A +1

I=

1 + 1/ 1

1 1/

1 1/0 0

and thus we find the corresponding eigenvector

1/1

.

(e) Find an invertible matrixPsuch thatA= P DP1 whereDis a diagonal matrix.

Proof. By Theorem 5 of Section 5.3, we must haveA= P DP1 where

D=

00 1/

, P =

1/1 1

.

(f) First give an explicit formula forDn and use this to give an explicit formula forAn.


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Proof. Powers of a diagonal matrix are easy to compute explicitly:

Dn =

n 00 (1/)n

.

Then we use the formulaA

n

=P D

n

P

1

to explicitly computeA

n

as follows:

An =

1/1 1

n 00 (1/)n

1

+ 1/

1 1/1

which boils down to

An = 1

+ 1/

n+1 (1/)n+1 n (1/)n

n (1/)n n1 (1/)n1

.

(g) Using parts (b) and (f), give an explicit formula for thenth Lucas number!

Proof. Using the formulas from parts (b) and (f), we obtain

Ln+1

Ln

= An

12

=

1

+ 1/

n+1 (1/)n+1 n (1/)n

n (1/)n n1 (1/)n1

12

= 1

+ 1/

n+1 (1/)n+1 + 2n 2(1/)nn (1/)n + 2n1 2(1/)n1

=

n+1 + (1/)n+1

n + (1/)n

.

(For the last simplification step, it really helps to notice that + 1/ = 1 + 2/ = 2 1.)Therefore, we finally obtain the remarkable formula

Ln= n + (1/)n.

proofs10 sol

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