# Homework Help: Find basis for vector space consisting of linear transformations

1. Dec 8, 2008

### p3forlife

1. The problem statement, all variables and given/known data
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) }

2. Relevant equations
L(W) is the set of linear transformations from W to W

3. 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.

2. Dec 8, 2008

### tiny-tim

Hi p3forlife!

Hint: what is the matrix for a general T?

what is the restriction on the matrix if T(x1) + T(x2) = T(x4)?

3. Dec 8, 2008

### p3forlife

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. Dec 8, 2008

### tiny-tim

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)?

5. Dec 8, 2008

### jambaugh

It may also help to use the linearity of your transformation $$T$$. You should be able to rewrite your condition as $T(u)=0$ for some specific vector $u$.

I am also not so sure of your dimension argument. Doesn't $$L(W)$$ 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 $$W \to$$ 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. Dec 8, 2008

### p3forlife

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)} ?

7. Dec 8, 2008

### tiny-tim

Hi p3forlife!

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)?

8. Dec 8, 2008

### p3forlife

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. Dec 8, 2008

### tiny-tim

ah, no …
T (a typical member of L(W)) is the transformation.

10. Dec 8, 2008

### p3forlife

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. Dec 9, 2008

### tiny-tim

Yes, but you haven't said why it's 4.

You can't put x1 etc in your matrix …

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

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)?

12. Dec 9, 2008

### Vid

L(W) has the same dimension as W.

Hint: A transformation can be completely determined by what it does to the basis vectors.

13. Dec 10, 2008

### jambaugh

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 $$L(W)=W\otimes W^*$$ which has the square of the dimension of W.