# Thread: Standard matrix of T

1. ## Standard matrix of T

Assume that R2>R2 is a linear transformation such that
T|1| =|1|
|2| |-3|

and
T|3| =|2|
|-1| |1|
Find the standard matrix of T.

Ok, so I get that |1,2|=e1+2e2 and that |3,-1|=3e1-e2, but what do I do from there on? It shows in the solutions that
e1=1/7|1,2| + 2/7|3, -1|
e2=3/7|1,2|-1/7|3,-1|

I understand that the are using the same thing from e1+2e2 and 3e1-e2, but where is the divisor of 7 coming from?

Sorry if it's formatted badly, I can scan the original page? Thanks

2. Okay so you know that

$T(e_1+2e_2)=e_1-3e_2$ and
$T(3e_1-e_2)=2e_1+e_2$

Now use the linearity of the operator to get

$T(e_1+2e_2)=T(e_1)+2T(e_2)=e_1-3e_2$ and
$T(3e_1-e_2)=3T(e_1)-T(e_2)=2e_1+e2$

This gives a system of equations that you can solve for $T(e_1),T(e_2)$

After you solve for them you should get

$T(e_1)=\frac{1}{7}(5e_1-e_2); T(e_2)=\frac{1}{7}(e_1-10e_2)$

These are the columns of the matrix representation of the transform.

$\begin{bmatrix} \frac{5}{7} & \frac{-1}{7} \\ \frac{1}{7} & \frac{-10}{7}\end{bmatrix}=\frac{1}{7}\begin{bmatrix} 5 & -1\\ 1 & -10\end{bmatrix}$

3. Thanks for the reply but I'm still confused. I don't get where you said it gives a system of equations for T(e1) and T(e2). Where would I get that from?

is it just
1 -1
2 1
Which I'm pretty sure is wrong... could you maybe put in the step of where you got the system of equations, and where from?

Thanks a lot

4. Originally Posted by shibble
Assume that R2>R2 is a linear transformation such that
T|1| =|1|
|2| |-3|

and
T|3| =|2|
|-1| |1|
Find the standard matrix of T.

Ok, so I get that |1,2|=e1+2e2 and that |3,-1|=3e1-e2, but what do I do from there on? It shows in the solutions that
e1=1/7|1,2| + 2/7|3, -1|
e2=3/7|1,2|-1/7|3,-1|

I understand that the are using the same thing from e1+2e2 and 3e1-e2, but where is the divisor of 7 coming from?

Sorry if it's formatted badly, I can scan the original page? Thanks
Let's take the following as given:
T is completely determined once the images of any basis for R^2 are known. *

With * in mind, I think I would read the initial statement, i.e.,
that T(1,2) = (1,-3) and T(3,-1) = (2,1), as telling me that
with respect to the basis B = {(1,2), (3,-1)} T is uniquely determined.

In this context the matrix $\begin{bmatrix}5/7 & 1/7 \\ -1/7 & -10/7\end{bmatrix}$ is the matrix representation of T
relative to basis B. Call it M'.

But, the problem asks you to find the "standard matrix" of T.
I read that as meaning the matrix representation of T relative to
the standard (or natural) basis E = {(1,0),(0,1)}.

So I think there's more work to do.

Can you say what the matrix $\begin{bmatrix}1/7 & 3/7 \\ 2/7 & -1/7\end{bmatrix}$ might represent?
Maybe the matrix of transition from E to B?
What about $\begin{bmatrix}1 & 3 \\ 2 & -1\end{bmatrix}$?

5. Originally Posted by shibble
Thanks for the reply but I'm still confused. I don't get where you said it gives a system of equations for T(e1) and T(e2). Where would I get that from?

is it just
1 -1
2 1
Which I'm pretty sure is wrong... could you maybe put in the step of where you got the system of equations, and where from?

Thanks a lot
$T(e_1+2e_2)=T(e_1)+2T(e_2)=e_1-3e_2$ and
$T(3e_1-e_2)=3T(e_1)-T(e_2)=2e_1+e2$

This is the system of equations or more explicitly

$T(e_1)+2T(e_2)=e_1-3e_2$ and
$3T(e_1)-T(e_2)=2e_1+e2$

Solving this gives you the transform of the standard basis.

6. Originally Posted by TheEmptySet
$T(e_1+2e_2)=T(e_1)+2T(e_2)=e_1-3e_2$ and
$T(3e_1-e_2)=3T(e_1)-T(e_2)=2e_1+e2$

This is the system of equations or more explicitly

$T(e_1)+2T(e_2)=e_1-3e_2$ and
$3T(e_1)-T(e_2)=2e_1+e2$

Solving this gives you the transform of the standard basis.
Hi, I don't know if its because it's late, or because I'm just really stupid, but I'm still really lost.

To solve, do you mean in this fashion:

$T(e_1)+2T(e_2)=e_1-3e_2$
$3T(e_1)-T(e_2)=2e_1+e2$

$\begin{bmatrix} 1 & -3 \\ 2 & 1\end{bmatrix}$

>>
$\begin{bmatrix} 1 & -3 \\ 0 & 7\end{bmatrix}$

so

$T(e_1)+2T(e_2)=-3$
$3T(e_1)-T(e_2)=7$

Just a wild guess here

I completely get this:

$T(e_1)+2T(e_2)=e_1-3e_2$ and
$3T(e_1)-T(e_2)=2e_1+e2$

and I also get that you have to use

$T(e_1)+2T(e_2)$ and
$3T(e_1)-T(e_2)$
where the e1 and e2 are (1,2) and (3,-1) respectively, but how do you find out what T is for each?
From what I'm looking at, the multiple stays the same, but there is a divisor for each T, in this case 7.
How do I solve for that T? Hopefully I'm being clear enough

EDIT I GOT IT, I guess I just needed some sleep Thanks guys, makes sense now