Hw

HW 6 page 86 #7

φ : L(V, V ) → L(W, W )φ(T ) = U T U −1 φ(T1 + T2 ) = U (T1 + T2 )U −1 = U T1 U −1 + U T2 U −1 as we want to show

is an isomorphism composition of linear

transformations is distributive.

φ(cT ) = U (cT )U −1 = (U (cT ))U −1 = (cU )T U −1 (as U is linear) = cφ(T ) So φ is linear −1 Now, dene ψ : L(W, W ) → L(V, V ), ψ(T ) = U TU ψ is linear by a similar argument. φ ◦ ψ(T ) = φ(U −1 T U ) = U (U −1 T U )U −1 = T , So φ ◦ ψ = IdL(W,W ) −1 Similarly, ψ ◦ φ(T)=ψ (UTU )=T=IdL(V,V ) , Hence φand ψ are inverses of each Also,

other As

φhas

an inverse,

φ

is a isomorphism.

Note: We could have also done this problem by checking that

φ

is injective or

surjective (Actually checking one is sucient as both L(V,V) and L(W,W) have

2

2

same dimension (dim V) =(dim W) ). To check surjectivity, for instance, we

φ(U−1 T'U)=

would need to take T'∈L(W,W) and say that

T'.

HW 7 #2 a) Show rank(S◦T)≤rank(S) and rank(S◦T)≤rank(T)}where S: W→X, T : V→W are linear transformations. let W' =T(V) - the range of T. W' is a subspace of W. similarly, let X' = S(W) - the range of S. X' is a subspace of X. also, let X=S◦T(V) - the range of S◦T. X is a subspace of X. by denition, rank(T) = dim W' , rank(S) = dim X' , rank(S◦T)= dim X Any vector in a in X is of the form S◦T(v) for some vector v in V. a = S◦T(v)=S(T(v)). so a is in the range of S, and hence a

∈X'.

So, any vector in X is in X' and hence, X

⊆

X', dim(X)≤dim(X')

rank(S)≤rank(S◦T) Also, since T(v)

∈W',

a=S(T(V)) is in the range of T|W 0 (T|W 0 : W'→X is the

restriction of T to W') X ⊆T(W'), So, dim X

≤dim

W'

rank(S◦T)≤rank T Hence, rank (S◦T) b) T

n+1

=T

n

◦

≤min{rank(S),

rank(T)}.

T, from part a) we get, rank T

n+1

≤Tn

hence rn+1 ≤rn as the sequence r1 ,r2 ,r3 ,... starts from some nite number and is never negative, it can decrease only nitely many times. So, it eventually becomes constant. #4 a) Choose the real vector space V=R, any non-zero real number r, gives a basis for V consisting of the single vector r.

1

B=(r) is a basis for V, now V*={linear functions from

∗

and has the dual basis B ={δr } where

δr

R to R}is one-dimensional

is the linear map which takes r to 1,

δr (x) = xr 1 1 Now since φV,B (ar)=aδr , we have φV,B (1)= δr = 2 δ1 r r ∗ So the map φV,B sends 1 to dierent elements in V for dierent

So,

values of r. So,

it depends on the basis. b) let B=(e1 , e2 , ..., en ) be a basis for V, a vector space over the eld F.

∗

∗

∗

B*=(e1 , e2 , ..., en ) be the dual basis to B for V*

∗∗ ∗∗ ∗∗ and B**=(e1 , e2 , ..., en ) be the dual basis to B* for V**. ∗ ∗ ∗∗ ∗∗ Now since φV,B (ei ) = ei and φV ∗ ,B ∗ (ei ) = ei , we have ψV,B (ei ) = ei ∗∗ ev(ei ), ei are both maps from V* to F and both of them take take ei to 1 and

ej (j 6= i)

to 0,

∗

∗

as ev(ei )(ej )=ej (e

i

) = δij (1

if i=j, 0 otherwise).

hence they must be the same map from V* to F. So

ψV,B (ei ) = e∗∗ i = ev(ei ).

As

ψV,B

and ev are equal on a basis for V and are

linear, they must be equal on all of V. HW8

R3 , W = R, T (x, y, z) = x + y + z T ={(x, y, z) : x + y + z = 0 x, y, z ∈ R} = {(x, y, −x − y) : x, y ∈ R}

4b) V = N=ker

dim N =2 now let v∈V, v=(a,b,c) the coset v+N consists of all vectors of the form (a+x, b+y, c+z) where x,y,z are real numbers such that x+y+z=0. T(a+x,b+y,c+z)=a+x+b+y+c+z=a+b+c. So T maps every vector in the coset v+W to the same element. Also, if another vector w=(p,q,r) has the sum (i.e. p+q+r=a+b+c), then it lies in v+W as (p,q,r)=(a,b,c)+(p-a,q-b,r-c) and (p-a,q-b,r-c)∈N. So T gives a bijection between the set of cosets and

R.

The dimension of V/W= dim R = 1 = dim V - dim N. 5 b) as T is surjective range(T)=W, rank T =dim W

t

rank T =rank T (you can use this fact directly, but a simple proof would be the following :

t

if A , B are the the associated matrices for T and T wrt to a basis B for V and

t the dual basis B* for V*, then B=A t = row rank of B = col rank of B t = rank T ) t ∗ t ∗ ∗ Nullity T =dim W - rank T = dim W - rank T = dim W - dim W =0 So, rank T =col rank of A = row rank of A

So, T is injective as the dimension of the kernel is 0. c) as T is injective, nullity T =0, rank T = dim V - nullity T =dim V

t

So, rank T

∗

= rank T = dim V= dim V

∗ ∗ range T which is a subspace of V has the same dimension as V . Hence range t ∗ T =V t

2

Hence, T

t

is surjective.

3