Math Is Fun Forum

corvax235

Member

Registered: 2007-05-05

Posts: 7

Linear transformation problem

Consider M_22 with the basis

{

,

,

,

}

and consider R^4 with the basis

.  Find the matrix which corresponds to the linear transformation L with respect to these two bases.

Last edited by corvax235 (2007-05-06 16:43:56)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear transformation problem

This is a fairly standard problem, what part are you having trouble on?  Just getting started?  Or is there something specific you don't get?

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

corvax235

Member

Registered: 2007-05-05

Posts: 7

Re: Linear transformation problem

I'm having trouble getting started.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear transformation problem

You should have been given a linear transformation L.  I'm going to assume since it wasn't given, that we can define it ourselves.  I'm almost certain you can't, but the problem is done in basically the same way.

To make the problem simple for me, I'm going to define L as:

L(a) = (7, 0, -3, 1)
L(b) = (2, -1, -2, 2)
L(c) = (-2, 0, 1, 0)
L(d) = (5, 17, 0, 0)

Where "a" is the first matrix in your basis, and so on.  Now if you don't see how this makes a full linear transformation, don't worry about it.  But trust me, it does.

Now I have a mapping which you are supposed to be given.  So to find the matrix, we first take each basis vector, transform it and it write each as a linear combination of the other basis.  For me, it's:

L(a) = (7, 0, -3, 1) = 1*(7, 0, -3, 1) + 0 * (2, -1, -2, 2) + 0 * (-2, 0, 1, 0) + 0 * (5, 17, 0, 0)
L(b) = (2, -1, -2, 2) = 0*(7, 0, -3, 1) + 1 * (2, -1, -2, 2) + 0 * (-2, 0, 1, 0) + 0 * (5, 17, 0, 0)
L(c) = (-2, 0, 1, 0) = 0*(7, 0, -3, 1) + 0 * (2, -1, -2, 2) + 1 * (-2, 0, 1, 0) + 0 * (5, 17, 0, 0)
L(d) = (5, 17, 0, 0) = 0*(7, 0, -3, 1) + 0 * (2, -1, -2, 2) + 0 * (-2, 0, 1, 0) + 1 * (5, 17, 0, 0)

Now we take these scalars (the numbers in bold), and each row makes a column in our transformation matrix:

1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear transformation problem

It's very important that you notice that you take the rows and write them as columns.  Unfortunately, in my example, it turned out to work the same no matter what way you do it.  But you write the ROWS as COLUMNS.  Everyone forgets that...

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

corvax235

Member

Registered: 2007-05-05

Posts: 7

Re: Linear transformation problem

Ah sorry, I forgot to include the linear transformation L

L of
(a  b
c  d)
=

(-10a - c + 2d, 5a + b - d, 2c + d, 9a + 17b + 2d)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear transformation problem

So now just take that transformation and do all the steps that I did with it.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

corvax235

Member

Registered: 2007-05-05

Posts: 7

Re: Linear transformation problem

Sorry I'm confused - how does
{

,

,

,

} fit into this?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear transformation problem

I just used letters in place of the matrix:

{a, b, c, d}

So calculate L(a), L(b), L(c), and L(d), as I did.  Your numbers will be a little more "interesting".

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

corvax235

Member

Registered: 2007-05-05

Posts: 7

Re: Linear transformation problem

L is:

L(

) = (-10, 5, 0, 9)
L(

) = (0, 1, 0, 17)
L(

) = (-1, 0, 2, 0)
L(

) = (2, -1, 1, 2)

L(

) = (-10, 5, 0, 9)  = x*(7, 0, -3, 1) + x * (2, -1, -2, 2) + x * (-2, 0, 1, 0) + x * (5, 17, 0, 0)
L(

) = (0, 1, 0, 17) = x*(7, 0, -3, 1) + x * (2, -1, -2, 2) + x * (-2, 0, 1, 0) + x * (5, 17, 0, 0)
L(

) = (-1, 0, 2, 0) = x*(7, 0, -3, 1) + x * (2, -1, -2, 2) + x * (-2, 0, 1, 0) + x * (5, 17, 0, 0)
L(

) = (2, -1, 1, 2) = x*(7, 0, -3, 1) + x * (2, -1, -2, 2) + x * (-2, 0, 1, 0) + x * (5, 17, 0, 0)

Last edited by corvax235 (2007-05-07 14:43:41)

corvax235

Member

Registered: 2007-05-05

Posts: 7

Re: Linear transformation problem

You know where I have all the x's filled in?  How do I know what numbers to put in place of those?

HallsofIvy

Guest

Re: Linear transformation problem

First replace those "x"s by DIFFERENT letters.  Then write out the equations you get for the individual components and solve those equations!

corvax235

Member

Registered: 2007-05-05

Posts: 7

Re: Linear transformation problem

Is there a way I can just use Mathematica to figure out the matrix?