# Vector Spaces and linear transformation

• May 12th 2012, 04:28 AM
tommyp123
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.
• May 12th 2012, 08:27 AM
Deveno
Re: Vector Spaces and linear transformation
(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:

$\begin{bmatrix}3&1&2&-1\\2&4&5&-1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix}$

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

row-reduction gives us:

$\begin{bmatrix}1&0&3/10&-3/10\\0&1&11/10&-1/10 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix}$

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).
• Apr 22nd 2013, 04:57 PM
jordan12345
Re: Vector Spaces and linear transformation
i think should be UnW is also a subspace of V . So how should we solve this question.
• Apr 22nd 2013, 05:19 PM
HallsofIvy
Re: Vector Spaces and linear transformation
Let x and y be vectors in $U\cap W$. Show that x+ y is also in $U\cap W$ and ax, for any number a, is in $U\cap W$.