Problem Set on Vector Spaces and Linear Transformations in Algebra

This assignment presents a series of solved problems in linear algebra, focusing on concepts such as linear maps, vector spaces, and linear transformations. It includes proofs and explanations related to the properties of linear maps between vector spaces, the span of vectors, and the kernel of transformations. Specific problems address topics like determining the null space of composite functions, proving that a set is a subspace, expressing vectors as linear combinations, and analyzing the image of geometric shapes under linear transformations. The solutions provide detailed steps and reasoning, making it a valuable resource for understanding these fundamental concepts in linear algebra. Desklib offers this and many other solved assignments to aid students in their studies.

Solution
Q6) let f:V → R , g :V → R are two linear maps from V to V
Such that
gof =R
gof ∗v ¿=0 for all v
g ( f ( v ) )=0 for all v
That is every vector in range space of f is a null space of gof
f ⊆ Null space of g
Hence
F is linear
Q7) let U =(v ∈ V ∨v=u−w)
Need to prove that U is a subset of V
Clearly U is a non−empty
Let a , b ∈ F∧v1, v2∈U
Then v1 = u1 – w1 and v2 = u2 – w2
Then
av1 + b v2 = a ¿1 – w1) + b ¿2 – w2)
= ¿1 –b u2) - ¿1 – bw2)
∈U ¿1 – bu2∈W and aw1 – bw2∈V as V and W are vector space)
Thus
av1 – bv2∈U for any a , b ∈ F∧v1, v2 ∈U
Q9) if V is a dimensional⊂of R
L :V → U
Where V ⊆ R is a⊂dimension one the { V } is a basic basis for the vector space V ⊆ R

Consider
t ∈ R
Since { V } is abasis for V ⊆ R any point t ∈V can be written as alinear combination of
the basis vector , Thus we must have t αV for some scalarw ∈ V .
Therefore
α ( t ) =α( v )
But since L ( v )={ L ( t ) :t ∈ v } we have
L ( v )={αL ( V ) :α ∈W }
Hence we can say that {L( v)}spans the entire L(v )
We now have 2 cases to consider L ( v ) =0∧L ( v ) ≠ 0
In the first case L ( v )=0=¿ L ( v )={0 }
This is a single point
in the second case
α ( v ) ≠ 0=¿ L ( v ) {αL ( v ) }
This is through a collection of points (x, y)
Satisfy if L( v) x
Where x = α=¿ y =L(v )
Hence L ( v ) is anologous¿ the line(one dimensional vector space )
so if v is one dimensional α ( v ) is anologious ¿ theline (one dimensional vector space)
L ( v ) can be a point of a plane
Q10) given v1, v2 are vectors in V and v3 = v1 + v2
Let
F ∈ span ¿1, v2 , v3)
The scalar a, b and c
F = av1+ bv2 + c v3)

F = (a + c)v1 + (b + c) v2
F∈ span ¿1, v2 )
Hence
F( v1), F( v2) ⊆ span ¿1, v2)
Q11)
L(0, 0) = (0, 0)
L ( 1, 0 ) = ( 1 ,2 )
L(1, 1) = (1 +2. 2 + 4) = (3, 6)
L ( 0 ,1 ) = ( 2 , 4 )
Determinant of the given matrix = 4 – 4 = 0
The image of the square is degenerated to parallelogram that is the line with the vertices
(0, 0), (3, 0), (0, 0) and (3, 1)
Q12)
For no c, A = cB => A and B vectors are not parallel also from non- parallel vector we define a plane that
is any vector lying in the plane that can be presented as linear combination of the 2 vectors how we are
given
T ¿1) = A and T ¿2) = B
E1 cannot be parallel to E2 and if it is so T being linear map say ¿1) =c ¿2)
A=T ( E 1 )=T ( E 2 ) =cT ¿2) = cB which is not possible
We have 4 possible vectors of which one is (0,0) and the rest 3 shall be transformed.
The transformation shall be a2A + b2B and a3A + b3B respectively given E1, E2

Q13)
Ker{B} ≠ {0}
Let dim(ker(B) = u > 0
And by rank nullity theorem
dim(R(B)) = dim A+dim (ker ( B ) )
Where
v is not invertible
Let dim ( A ) =t ≥ u
Therefore
dim ¿
let A = {w1 …wm} and B = {wm+1 …wn} be basis for
ker ( B )∧R ( B ) respectively
Then
A B is basis of v
Define c : A → A1 as c ¿1w1 … cnwn) - c1w1 + … cnwn
Then c islinear transformation∧c ¿1) = w1 ≠ 0:c is a zero transformation
For any A ∈ A1
B ( A )=α m+1wm+1 + … αnwn [B(A) R( B)¿
c B ( A ) =c ( B ( A ) ) = A ¿m+1wm+1 + … αnwn ) = 0
Therefore,
c B is a zero transformation