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Homework assignment no II

Mikoaj Paraniak

April 7, 2013

Excercise 1

Let A GL5(R

)

A=

1 0 0 0 01 1 0 0 00 0 2 0 10 0 0 2 00 0 0 1 2

In order to compute desired quantities we first find the jordan normal form of A.det(A I5) is easy to find. The result can be verified by the reader on the fly:

det(A I5) = (1 )2(2 )3 (1)

Thus we obtain two eigenvalues:

1= 1

2= 2

Let us find the chain generated by 1We see thatker(A1I5) = ker(A I5). Again,finding the base ofker(A I5) is not chalenging, thus the exact steps are omitted. Weobtain:

ker(A I5) =span

01000

We also compute ker(A I5)2

.

1

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(A I5)2 =

0 0 0 0 00 0 0 0 00 0 1 1 20 0 0 1 0

0 0 0 2 1

Thus

ker(A I5)2 =span{m1,m2}= span

10000

,

01000

We see that m1 is not in ker(A I5). We solve the equation (A I5)m1= m1 (then

(A I5)m1= 0), and obtain

m1= (0, 1, 0, 0, 0)

t

Since geometric multiplicity need of1 not exceed its algebraic multiplicity the chaingenerated by1 is{m

1,m1}.

Now we consider the chain generated by 2. Since the geometric multiplicity of1wasfound to be 2, the geometric multiplicity of2must be equal to 3. We findker(A2I5) =

ker(A 2I5) to be:

ker(A 2I5) =span

00100

And ker(A 2I5)2:

ker(A 2I5) ={n1,n2}= span

00001

,

00010

We see that none of n1, n2 is in ker(A 2I5). Thus we can easily find the baseker(A 2I5)

3 by arbitrarily choosing n1:

ker(A 2I5)n1= n1= (0, 0,1, 0, 0)t

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Thus we have three linearly independent vectors n1,n1,n2, all contained in ker(A 2I5)

3, therefore they form the base ofker(A 2I5)3. We found the jordan normal form of

A, A= P J P1

A=

0 1 0 0 01 0 0 0 00 0 1 0 00 0 0 0 10 0 0 1 0

1 1 0 0 00 1 0 0 00 0 2 1 00 0 0 2 10 0 0 0 2

0 1 0 0 01 0 0 0 00 0 1 0 00 0 0 0 10 0 0 1 0

Now we can easily find exp A. We see that

exp

1 10 1

=

=e exp

0 10 0

=e(I+

0 10 0

+1

2

0 10 0

2) =

=e

1 10 1

also

exp

2 1 00 2 1

0 0 2

=

=e2 exp

0 1 00 0 1

0 0 0

=

=e2(I+

0 1 00 0 1

0 0 0

+1

2

0 0 10 0 0

0 0 0

) =

= e2(

1 1 12

0 1 1

0 0 1

)

finally

eJ =

e e 0 0 00 e 0 0 00 0 e2 e2 1

2e2

0 0 0 e2 e2

0 0 0 0 e2

And

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exp A= P eJP1 =

e 0 0 0 0e e 0 0 00 0 e2 1

2e2 e2

0 0 0 e2 0

0 0 0 e2

e

2

sin A= P(sin J)P1

sin J=

sin 1 cos 1 0 0 00 sin 1 0 0 00 0 sin 2 cos 2 1

2sin 2

0 0 0 sin 2 cos 20 0 0 0 sin 2

sin A= P(sin J)P1 =

sin 1 0 0 0 0cos 1 sin 1 0 0 0

0 0 sin 2 12sin 2 cos2

0 0 0 sin 2 00 0 0 cos 2 sin 2

Excercise 2

Let B ={e1, e2, e3, e4, e5}

B =

01111

,

10111

,

11011

,

11101

,

11110

be a base of a vector space C5. LetB ={ei}be a dual base ofC5. From definition ofa dual base, B ={ei}is the base of a vector space C5

=Map(C5,C)such that following

condition is satisfied: For every ei B there is ei 5

k=1

ckek

=ciwhere ck C. Taking

ci= 1, ck=i= 0 we obtain:

eiej =ij (2)

Since ei is a linear map, and dimC5

=dim C5 (because dimC5 is finite), we can writethe matrix ofei as:

ei =

ei1, ei2, ei3, ei4, ei5

(3)

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where eik C. Thus, from(1) and (2) we obtain a system of equations for e1, which

we write in a matrix form:

0 1 1 1 11 0 1 1 11 1 0 1 11 1 1 0 11 1 1 1 0

=

10000

The solution is easy to find, and the exact steps are not included here. We get

e1 =

1

4

3, 1, 1, 1

Indeed, let S=

5k=1

ck, and C5

l=5

k=1 ckek Then we have:

e1l=

1

4

3, 1, 1, 1, 1

S c1S c2S c3S c4S c5

=

1

4(3S+ 3c1+ 4S (S c1)) =

1

4(4c1) =c1

Derivation of remaining dual base vectors is analogous. We have:

e2 =

1

4

1, 3, 1, 1, 1

e3 =

1

4

1, 1, 3, 1, 1

e4 =

1

4

1, 1, 1, 3, 1

e5 =

1

4 1, 1, 1, 1, 3and

B ={e1, e2, e3, e4, e5}

as was to be found.