• Support PF! Buy your school textbooks, materials and every day products Here!

Find basis for vector space consisting of linear transformations

  • Thread starter p3forlife
  • Start date
  • #1
20
0

Homework Statement


Find a basis for V.
Let W be a vector space of dimension 4. Let beta = {x1, x2, x3, x4 } be an ordered basis for W. Let V = {T in L(W) | T(x1) + T(x2) = T(x4) }


Homework Equations


L(W) is the set of linear transformations from W to W


The Attempt at a Solution


We know that V has dimension 3, since T(x4) can be expressed in terms of T(x1) and T(x2). So V has the general form of {T(x1), T(x2), T(x3), T(x1) + T(x2)}
But I'm lost as to how I can find a basis for V when nothing is said explicitly.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi p3forlife! :smile:

Hint: what is the matrix for a general T?

what is the restriction on the matrix if T(x1) + T(x2) = T(x4)? :wink:
 
  • #3
20
0
So the matrix for a general T will look like:

[T] = [a11 a12 a13 a14]
[a21 a22 a23 a24]
[a31 a32 a33 a34]
[a41 a42 a43 a44]

where a11 + a12 = a14
a21 + a22 = a24
a31 + a32 = a34
a41 + a42 = a44

but i can't get any farther than this unfortunately.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
So the matrix for a general T will look like:

[T] = [a11 a12 a13 a14]
[a21 a22 a23 a24]
[a31 a32 a33 a34]
[a41 a42 a43 a44]

ok, let's do this slowly and logically …

the basis for V must be smaller than any basis for L(W) …

can you find a basis for L(W)? :smile:
 
  • #5
jambaugh
Science Advisor
Insights Author
Gold Member
2,218
250
It may also help to use the linearity of your transformation [tex]T[/tex]. You should be able to rewrite your condition as [itex] T(u)=0[/itex] for some specific vector [itex]u[/itex].

I am also not so sure of your dimension argument. Doesn't [tex]L(W)[/tex] start with dimension 16?

Remember that (when acting on the left on a column vector) each row of a matrix is a linear functional mapping [tex]W \to [/tex] the real numbers.

You can start with that simpler problem of finding a basis for the space of all linear functionals satisfying the given condition. Its answer should be very helpful in this problem.
 
  • #6
20
0
Okay, so finding a basis for L(W)...
Since beta = { x1, x2, x3, x4} is a basis for W, if you do a linear transformation from W to W, the basis should be { T(x1), T(x2), T(x3), T(x4)} ?

Sorry...I'm striking a blank about this problem.
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
251
Okay, so finding a basis for L(W)...
Since beta = { x1, x2, x3, x4} is a basis for W, if you do a linear transformation from W to W, the basis should be { T(x1), T(x2), T(x3), T(x4)} ?

Sorry...I'm striking a blank about this problem.
Hi p3forlife! :smile:

I don't think you understand what the elements of L(W) are …

and a basis isn't attached to a particular T … it has to be the same basis for all T in L(W).

How would you define a particular element of L(W)? :smile:
 
  • #8
20
0
So L(W) means you take an x in W, you apply a transformation, then you get T(x), where the set of all T(x) is the range of W.
 
  • #9
tiny-tim
Science Advisor
Homework Helper
25,832
251
So L(W) means you take an x in W, you apply a transformation, then you get T(x), where the set of all T(x) is the range of W.
ah, no …
L(W) is the set of linear transformations from W to W
T (a typical member of L(W)) is the transformation. :smile:
 
  • #10
20
0
Argh....sorry this is taking so long :S

So since L(W) is the set of all linear transformations from W to W, it means that V is a subset of a W, since there is the restriction of T(x1) + T(x2) = T(x4)
In L(W), we start off with 16 dimensions, since each of T(xi) for i = 1, 2, 3, 4 has 4 elements, and 4 times 4 is 16. But, in V, since T(x1) + T(x2) = T(x4), we can take off 4 dimensions from 16, giving us 12 dimensions.

So in order to satisfy T(x1) + T(x2) = T(x4),
choose T(x1) = 1, T(x2) = T(x3) = 0, T(x4) = 1. So then the basis for V has to be in terms of x1, x2, x3, x4, so we have the matrix:

x1 0 0 x1
x2 0 0 x2
x3 0 0 x3
x4 0 0 x4
0 x1 0 x1
0 x2 0 x2
0 x3 0 x3
0 x4 0 x4
0 0 x1 0
0 0 x2 0
0 0 x3 0
0 0 x4 0

Is this what you mean by a matrix for a general T?
 
  • #11
tiny-tim
Science Advisor
Homework Helper
25,832
251
… But, in V, since T(x1) + T(x2) = T(x4), we can take off 4 dimensions from 16, giving us 12 dimensions.
Yes, but you haven't said why it's 4.

So in order to satisfy T(x1) + T(x2) = T(x4),
choose T(x1) = 1, T(x2) = T(x3) = 0, T(x4) = 1. So then the basis for V has to be in terms of x1, x2, x3, x4, so we have the matrix:

Is this what you mean by a matrix for a general T?
You can't put x1 etc in your matrix …

the elements of the matrix are numbers, and x1 is a vector. :frown:

A general matrix is

a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44

where the a's are all numbers,

and the vector x1 + x2, for example, is

1
1
0
0

So how would you write, using vectors and a matrix, T(x1) + T(x2) = T(x4)? :smile:
 
  • #12
Vid
401
0
L(W) has the same dimension as W.

Hint: A transformation can be completely determined by what it does to the basis vectors.
 
  • #13
jambaugh
Science Advisor
Insights Author
Gold Member
2,218
250
L(W) has the same dimension as W.

Hint: A transformation can be completely determined by what it does to the basis vectors.
The way he described it L(W) does not have the same dimension as W. L(W) is not the dual space of functionals W* (which does have same dimension as W if it's finite)
but rather [tex]L(W)=W\otimes W^*[/tex] which has the square of the dimension of W.
 

Related Threads on Find basis for vector space consisting of linear transformations

  • Last Post
Replies
11
Views
1K
Replies
11
Views
2K
Replies
2
Views
6K
Replies
8
Views
7K
Replies
0
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
0
Views
5K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
0
Views
1K
Top