Thread: Vector Spaces and linear transformation

(a) Suppose that
V is vector spaces over a field F and that U and W are subspaces of V .
Show that
U \W is also a subspace of V

(b) Define a real linear transformation
L1 : R4 -->R2 by

L1(x1, x2, x3, x4) = (3x1 + x2 + 2x3 − x4, 2x1 + 4x2 + 5x3 − x4)
and let
U1 denote the kernel of L1. Define a real linear transformation L2 : R4 --> R2 by
L
L2=(x1, x2, x3, x4) = (5x1 + 7x2 + 11x3 + 3x4, 2x1 + 6x2 + 9x3 + 4x4)
and let
U2 denote the kernel of L2. Construct bases for U1, U2, U1 nU2 and U1 + U2.

(a) what do you mean by U\W? if you mean the set difference, this cannot possibly be a subspace of V, since since {0} is in the intersection, and thus NOT in U\W.

on the other hand, if you mean the quotient of U by W, this isn't necessarily even defined, since W may not be a subspace of U. something is missing, here...

(b) well first we need to determine the two kernels (solve two homogeneous sets of linear equations). for example:

ker(L₁) = {(x₁,x₂,x₃,x₄) in R⁴ : L₁(x₁,x₂,x₃,x₄) = (0,0)}

this is equivalent to the homogeneous system of linear equations:

3x₁ + x₂ + 2x₃ - x₄ = 0
2x₁ + 4x₂ + 5x₃ - x₄ = 0

or the matrix equation:



it should be clear that this matrix has rank 2, so the kernel has dimension 2.

row-reduction gives us:



so if we set x₃ = s, x₄ = t, we get:

10x₁ = -3s + 3t
10x₂ = -11s + t

so the kernel consists of vectors of the form (-3s+3t,-11s+t,10s,10t) = s(-3,-11,10,0) + t(3,1,0,10), so {(-3,-11,10,0),(3,1,0,10)} is a basis for U₁.

i leave it to you to construct a basis for U₂, and of U₁∩U₂, U₁+U₂.

you may find it helpful to know that dim(U₁∩U₂) ≤ min(dim(U₁),dim(U₂)) and that:

dim(U₁+U₂) = dim(U₁) + dim(U₂) - dim(U₁∩U₂).

i think should be UnW is also a subspace of V . So how should we solve this question.

Let x and y be vectors in . Show that x+ y is also in and ax, for any number a, is in .