How to determine a formula for T using given linear transformations?

EV33
Messages
192
Reaction score
0

Homework Statement


Derive a formula for T.

T([1 1]^T)=[2 -1]^T and

T([1 -1]^T)=[0 3]^T

(...^T=transpose and T(...)=Linear Transformation

Homework Equations



T(c1v1+...+cnvn)=c1T(v1)+...+cnT(vn)

The Attempt at a Solution



The solutions manual's method and the method I am supposed to use for these problems.

Let u1=[1 1]^T and u2=[0 -1]^T. If x=[x1 x2]^T then x=[(x1+x2)/2]u1+[(x1-x2)/2]u2...

this is the full solution but I don't know what they are doing here. My question is where did the 1/2's come from?

It says to use the equation that have in the relavant equations part, but I don't see how that would get me where they are.

Using that I would get...

T(u1+u2)=T(u1)+T(u2)=a1u1+a2u2=a1[2 -1]^T + a2[0 3]^T

and I am stuck at this point and don't see a connection between what I have and what they have.
 
Physics news on Phys.org
You really want to know what T does to the standard basis {[1 0]^T, [0 1]^T}. As it turns out, [1 0]^T = 1/2 * [1 1]^T + 1/2* [1 -1]^T, and
[0 1]^T = 1/2 * [1 1]^T - 1/2* [1 -1]^T.
 
I don't be a pain here but I don't see the connection. The problem doesn't say anything about the standard basis.

So am I supposed to set up 2 matrices each time I have a problem like this and put e1 in the augmented side of 1, and e2 in the augmented side of the other?
 
If you know what T does to [1 0]T and [0 1]T, then you know what it does to any arbitrary vector [x y]T. The reason is that [x y]T = x* [1 0]T + y*[0 1]T.

So T([x y]T) = T(x* [1 0]T + y*[0 1]T) = x*T([1 0]T) + y*T([0 1]T).
 
Thank you for all the help so far Mark 44, but I am having trouble getting this.

I know what we are trying to do here. But this process makes no sense to me.I know we are just looking for an equation that we plug x from T(X) into to get T(x). I can do these easy problems obviously without doing this process, but I am guessing these can get a lot uglier so if someone could just walk me through a full problem I think that might really help me. If someone is up for that I can post a problem or if you have a good example problem that works too.

Let's try T([1 1]^T)=[1 2 1]^T
and T([1 -1]^T)=[0 2 2]^T

Let's call the vectors being transformed u1 and u2

x=a1u1+a2u2
T(x)=a1[1 2 1]^T+a2[0 2 2]^T

then I set up a matrix
1 1 1
1 -1 0

1 1 0
1 -1 1

then I get the x1 and x2 of each of those to be 1/2.

So a1=[(x1+x2)/2),(2x1+2x2)/2),(x1+x2)/2]^T
a2=[(x1+x2)/2),(x1-x2)/2),(x1-x2)/2)]^T

Is this right so far?

If so how do I finish the problem up?
 
EV33 said:
Thank you for all the help so far Mark 44, but I am having trouble getting this.

I know what we are trying to do here. But this process makes no sense to me.I know we are just looking for an equation that we plug x from T(X) into to get T(x).
No, that's not it. We aren't looking for an equation we can plug in. We don't know T(x). What we would like is some matrix A such that T(x) = Ax.

All we know are T(u) and T(v) for some arbitrary vectors u and v. We would like to know what T(x) is for any old vector x. We're trying to figure out what the transformation does to a basis, because if we know that, we know what it will do to any vector. This is what I said in post #4.
EV33 said:
I can do these easy problems obviously without doing this process, but I am guessing these can get a lot uglier so if someone could just walk me through a full problem I think that might really help me. If someone is up for that I can post a problem or if you have a good example problem that works too.

Let's try T([1 1]^T)=[1 2 1]^T
and T([1 -1]^T)=[0 2 2]^T

Let's call the vectors being transformed u1 and u2
OK. By inspection, I can see that (1/2)(u1 + u2) = [1 0]T, and also that (1/2)(u1 - u2) = [0 1]T. This shouldn't be too much of a surprise, since u1 and u2 are the same vectors as in the first problem.
EV33 said:
x=a1u1+a2u2
T(x)=a1[1 2 1]^T+a2[0 2 2]^T
No, you can't say that. You don't know what T(x) is - that's really what the problem wants you to find. Also, let's leave x out of things. Here's what we know.
T([1 0]T) = T((1/2)(u1 + u2)) = (1/2)T(u1) + (1/2)T(u2) = [1/2 2 3/2]T

Also, T([0 1]T) = T((1/2)(u1 - u2)) = (1/2)T(u1) - (1/2)T(u2) = [1/2 0 -1/2]T

In the work above, I used what you gave me for T(u1) and T(u2).

Now, we're ready to define T(x), where x is the vector [x1 x2]T.

T(x) = T([x1 x2]T) = T(x1*[1 0]T + x2*[0 1]T)
= x1*T([1 0]T) + x2*T([0 1]T)
= x1* (what?) + x2*(what?)

This defines T(x).

As a side note, T([1 0]T) and T([0 1]T) can be used to define the columns of a 3 x 2 matrix that I'll call A.

You can also define the transformation T(x) using the matrix:
T(x) = Ax

For this problem, A turns out to be
[1/2 1/2]
[2 0 ]
[3/2 -1/2]

EV33 said:
then I set up a matrix
1 1 1
1 -1 0

1 1 0
1 -1 1

then I get the x1 and x2 of each of those to be 1/2.

So a1=[(x1+x2)/2),(2x1+2x2)/2),(x1+x2)/2]^T
a2=[(x1+x2)/2),(x1-x2)/2),(x1-x2)/2)]^T

Is this right so far?

If so how do I finish the problem up?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top