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MA 106: Spring 2014: Tutorial Sheet 5

1. Find the best approximation ofv = [1, 3, 1, 5]t R41 from each of the following subspaces ofR41:W0 = {[0, 0, 0, 0]t}, W1 = L({[1, 0, 0, 0]t}), W2 = L({[1, 0, 0, 0]t, [1, 1, 0, 0]t}),W3 = L({[1, 0, 0, 0]t, [1, 1, 0, 0]t, [1, 1, 1, 0]t}) andW4 = L({[1, 0, 0, 0]t, [1, 1, 0, 0]t, [1, 1, 1, 0]t, [1, 1, 1, 1]t}).Also, find the distance ofv from each of the above best approximations.

Solution Orthonormalization of {(1, 0, 0, 0)t, (1, 1, 0, 0)t, (1, 1, 1, 0)t, (1, 1, 1, 1)t} by the G-S processgives w1 = (1, 0, 0, 0)

t, w2 = (0, 1, 0, 0)t, w3= (0, 0, 1, 0)

t, and w4 = (0, 0, 0, 1)t.

Letdj denote the distance ofv = (1, 3, 1, 5)t fromWj =L({w1, . . . , wj}), j = 1, . . . , 4.

HencePWj (v) =j

i=1wi, vwi andd2j =v

2 j

i=1 |wi, v|2. Thus

PW0(v) = (0, 0, 0, 0)t, d20 = v

2 = 36.

PW1(v) = (1, 0, 0, 0)t, d21= 36 1 = 35.

PW2(v) = (1, 3, 0, 0)t, d22= 36 1 9 = 26.

PW3(v) = (1, 3, 1, 0)t, d23= 36 1 9 1 = 25.

PW4(v) = (1, 3, 1, 5)t, d24= 36 1 9 1 25 = 0.

2. (a) Findx = [x1, x2]

t

R21

such that the straight liney = x1+ x2xbest fits the data points (1, 2),(0, 0), (1, 3) and (2, 5) in the least squares sense.

(b) Findx = [x1, x2, x3]t R31 such that the parabolay = x1 + x2x + x3x2 best fits the data points(1, 1), (1, 1), (2, 3) and (2, 2) in the least squares sense.

Solution (a) A =

1 11 01 11 2

, x=

x1x2

, b=

20

35

.

AtA=

4 22 6

, Atb=

615

.

The unique solution to the normal equations AtAx= Atb is x1 =3/10 and x2 =12/5. Thus theanswewr is y = (3/10) + (12/5)x.

(b)A =

1 1 11 1 11 2 41 2 4

, x=

x1x2

x3

, b=

1132

.

AtA=

4 0 100 10 0

10 0 34

, Atb=

72

22

.

The solution to the normal equations is x1 = 1/2, x2 = 1/5, x3 = 1/2. Hence the answer is y =1/2 + (1/5)x+ (1/2)x2.

3. Letp(t) = k0+k1t+ +kmtm be a polynomial with complex coefficients. For annncomplex matrixA, define then nmatrixp(A) = k0In+ k1A + + kmAm. Let1, . . . , n be then eigenvalues ofA

(with possible repetitions). Show that the eigenvalues ofp(A) are p(1), . . . , p(n).

Solution IfU is an upper triangular matrix with diagonal entries uii, then the diagonal entries ofUl

areulii. The result now follows from Schurs theorem.

4. Find the eigenvalues and their geometric as well as algebraic multiplicities of the following matrices.Are they diagonalizable? If so, find P such that P1APis diagonal in each case.

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(i)A :=

5 11 3

, (ii) A :=

3 2 1 00 1 0 10 2 1 00 0 0 1/2

.

Solution (i) The only eigenvalue ofA is 4 with algebraic multiplicity 2. Now

A 4I=

1 11 1

.

Clearly, nullity ofA is 1, so geometric multiplicity of the eigenvalue 4 is 1 < 2 = Algebraic multiplicity.Thus A is NOT diagonalizable.

(ii) We have

A tI=

3 t 2 1 00 1 t 0 10 2 1 t 00 0 0 1/2 t

.

By inspectiondet(A tI) = (3 t)(1/2 t)(1 t)(1 +t).

SoA has four distinct eigenvalues 3, 1/2, 1, 1 and is thus diagonalizable.

Here are the eigenvectors (upto sclar multiples):

(1, 0, 0, 0)t is an eigenvector with eigenvalue 3.

(3/2, 1, 1, 0)t is an eigenvector with eigenvalue 1.

(1, 0, 4, 0)t is an eigenvector with eigenvalue 1.

(8, 6, 8, 3)t is an eigenvector with eigenvalue 1/2.

Thus

P =

1 3/2 1 80 1 0 60 1 4 80 0 0 3

.

5. Find necessary and sufficient conditions on x,y, z for the matrix

2 x y0 1 z

0 0 2

to be diagonalizable.

SolutionDenote the matrix in question byA. The eigenvalues ofA are 1, 2, 2 and thusA is diagonal-izable iff dimN(A 2I) = 2. Now

A 2I=

0 x y0 1 z

0 0 0

.

So dimN(A 2I) = 2 iffxz +y = 0.

6. For a complex nn matrix A = [aij ], define

A1 = max n

i=1

|aij |: j = 1, . . . , n

and A2 = max

Ax: x Cn1 withx 1

.

For an eigenvalue ofA, show that || min{A1, A2}.

Solution Let be an eigenvalue ofA.

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Let a = A1. There is a nonzero row vector x = (x1, . . . , xn) such that xA = x. Put =max{|x1|, . . . , |xn|}. Note that >0.

Forj = 1, . . . , n, we have

xj =ni=1

aijxi.

Thus, for j = 1, . . . , n,

|||xj |= |ni=1

aijxi| ni=1

|aij ||xi| (ni=1

|aij |)a.

Choosing j0 such that |xj0 |= gives a, since >0.

Now let b = A2. Let x Cn1 satisfyAx= b, x 1. Then A xx=

bx b, sox 1 and

thusx= 1.

Let 0=y be a unit eigenvector with eigenvalue . Then

||= ||y=y=Ay b.

7. (Gershgorin) Let A = [aij] be an nn complex matrix. For i = 1, . . . , n, let Di denote the closeddisk in the complex plane with center aii and radius ri, where

ri = |ai1| + |ai2| + + |ain| |aii|.

Show that every eigenvalue ofA lies in D1 D2 Dn.

Solution Let be an eigenvalue ofA with eigenvectorx = (x1, . . . , xn)t, Then (A I)x= 0 and for

eachi = 1, . . . , nwe have

ai1x1+ +ai1,1xi1+ (aii )xi+ai+1,1xi+1+ +ainxn = 0.

Let|xk|= max{|x1|, . . . , |xn|}> 0. Then

akk =

ak1x1xk + +a

kn

xnxk a

kk

.

Taking absolute value on both sides we get

|akk | |ak1| + + |akn| |akk|= rk.

Thus Dk.

8. Show that any two nonzero 2 2 complex nilpotent matrices are similar.

Solution Check that any 2 2 complex nilpotent nonzero matrix is similar to

0 10 0

.

9. Let A be a complex nn matrix with eigenvalues 1, . . . , n. Fix an arbitrary ordering of the is.

Show that there exists an nn invertible P such that P1

AP is upper triangular and the diagonalentries ofP1AP(that is, the eigenvalues ofA) appear in the order fixed above.

Solution The proof of Schurs theorem given in the notes works in this slightly greater generality.

10. For an eigenvalue of a complex nn matrix A, define the generalized eigenspace correspondingto by

W = {x Fn : (A I)nx= 0}.

Show that the dimension ofW is equal to the algebraic multiplicitly of. (Hint: Use Problem 9.)

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SolutionLet the algebraic multiplicity of be k. List the eigenvalues ofA in such a way that the firstk are. Then, by Problem 9, A is similar to an upper triangular B whose first k diagonal entries areand the other diagonal entries are=. LetV denote the generalized eigenspace ofB correspondingto. Since A and B are similar we have dim V = dim W.

Now the first k rows and first k columns of B I are upper triangular with 0s on the diagonal.Moreover, the othern k diagonal entries ofB Iare= 0. It follows that V= L({e1, . . . , ek}) and

hence dim V= k.

11. LetA and B be 3 3 complex matrices each of which has only two distinct eigenvalues. Assume thatthese two eigenvalues are the same for A and B and denote them by and .

(a) If the algebraic multiplicities of and are the same for both A and B, mustA andB be similar?

(b) If the geometric multiplicities of and are the same for both A and B, must A and B besimilar?

(c) If the algebraic as well as the geometric multiplicities of and are the same for both A and B ,must A and B be similar?

(Hint: Use Problem 8.)

Solution a) NO. Take A = 1 0 0

0 1 00 0 2

and B =

1 1 00 1 00 0 2

.

b) NO. TakeA =

2 1 00 2 0

0 0 1

and B =

1 1 00 1 0

0 0 2

.

c) YES. Without loss of generality we may assume that algebraic multiplicity of is 1 and that ofis 2. Let k = geometric multiplicity of in A (or B ).

Case (i): k= 2. Then both A and B are diagonalizable and are clearly similar.

Case (ii): k = 1. List the eigenvalues ofA as ,, . By problem 9, A is similar to

A

=

0 0 0

.

Since =, A is similar to

A =

00 0

0 0

.

(reason: Take an eigenvectorv = (a1, a2, a3)t with eigenvalue. Thena3 = 0. Now, A is the matrixofTA wrt the ordered basis (e1, e2, v)). Note that = 0 as otherwisek = 2.

SimilarlyB is similar to

B =

00 00 0

,

with= 0.

Using Problem 8 we see that A andB are similar.

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