# Math Help - [SOLVED] Transformation

1. ## [SOLVED] Transformation

Let
$\{T: R^n \to R^n |T{\text{ is linear\} }}$

$\left[ {\begin{array}{c}
x \\
y \\
z \\
\end{array}} \right] = \left[ {\begin{array}{c}
{x + y} \\
z \\
\end{array}} \right]

$

2. You are way fancier with the LaTeX code than me, I apologize in advance.
Let $u=$
Let $v=$
Let $a\in \mathbb{R}$
You must show that (1) $T(u + v)=T(u)+T(v)$ and (2) T(au)=aT(u).

(1)
$T(u+v)=T()=$
$T(u) + T(v) = + = $

(2)
$T(au)=T()= =a=aT(u)$

3. Sorry but i still dont have any idea on how i can show $T$ as a matrix. thanks gamma for help in this and past posts.

4. Okay, so I think maybe the problem lies in understanding this concept of a linear transformation.

A linear transformation T is talking about functions from vector spaces. It is a property of a function that needs to be checked, one must check that T respects addition and scalar multiplication, the two operations that define a vector space. This explicitly means checking that T(u+v)=T(u)+T(v) and T(cu)=cT(u) for any arbitrary vectors u,v in your domain vector space, and c is just some scalar from your field, generally this will be the real numbers, as these are "real vector spaces."

In that problem you were asked to show that the function was a linear transformation, so you need only check those two things which I did for you. That is how you show a function is linear.

I put a lot of space here for a reason: to stress that matrices are different from linear transformations. A matrix is simply a convenient way to represent what a linear transformation does to a vector. Instead of having to plug in a vector to a function, you can simply take a matrix A representing your function and know that Av=T(v).

Now I dont know if your question tells you explicitly somewhere else to write the transformation as a matrix or not, because in your post the question says nothing about this, it says only check that it is linear.

But it is true that for a linear transformation you can build a matrix that will represent the transformation, and it is done as follows. You take your basis elements from the domain (I will assume the standard basis but it should be noted that this does not need to be the choice). Look at where the function sends these basis elements. Wherever it gets sent into the range you can write this as a linear combination of the basis elements (remember the order matters in the basis) this is the definition of a basis, it is linearly independent, spans and is unique. So you pick off the coefficients from this linear combination that gives you the image of your first domain basis vector. This is your first column. Now do the same for the second basis element, that is your second column. and so on.

Here is how you would do it for your example.

The first basis element is (1,0,0). This gets sent to [1+0,0]=[1,0] so your first column is 1,0.

The second basis element is (0,1,0) it gets sent to [0+1,0]=[1,0] so your second column is 1,0.

The third basis element is (0,0,1) it gets sent to [0+0,1]=[0,1] so your last column is 0,1.

Check that these dimensions make sense, you have a 2x3 matrix acting on a 3 dimensional vector and you are hoping to get a 2 dimensional vector out.
2x3 3x1 = 2x1 this is good.

Now the matrix A we got has first row 1,1,0 second row 0,0,1. On the same page?

Just to convince you that it worked. Go ahead and multiply A<x,y,z>, what do you know? You get exactly, <x+y,z>. Congratulations, you have just found the matrix representation of a linear transformation.

5. ## Sleep

It is 4am here, I gotta hit the sack. I hope I was explicit enough in the last post. It is a tough thing to wrap your head around, it just takes time and effort and experience playing around with it. I hope this makes sense to you, Ill check on it in the morning, good night.

6. thanks a lot for your help.

7. Originally Posted by Gamma
You are way fancier with the LaTeX code than me, I apologize in advance.
Let $u=$
Let $v=$
Let $a\in \mathbb{R}$
You must show that (1) $T(u + v)=T(u)+T(v)$ and (2) T(au)=aT(u).

(1)
$T(u+v)=T()=$
$T(u) + T(v) = + = $

(2)
$T(au)=T()= =a=aT(u)$

Am i on the right? This is what i did.
$T(u + v) = T(u) + T(v)$
$T(u + v) = T\left[ {\begin{array}{*{20}{c}}
{{x_1} + {x_2}} \\
{{y_1} + {y_2}} \\
{{z_1} + {z_2}} \\
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{x_1} + {y_1}} \\
{{z_1}} \\
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{x_2} + {y_2}} \\
{{z_2}} \\
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{x_3} + {y_3}} \\
{{z_3}} \\
\end{array}} \right]
$

$
T(u) + T(v) = \left[ {\begin{array}{*{20}{c}}
{{x_1} + {x_2}} \\
{{x_3}} \\
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{y_1} + {y_2}} \\
{{y_3}} \\
\end{array}} \right] = ?
$

I don't know what to do next or even if what i did is right? If i am wrong could you please tell me where i went wrong?
Thanks
KC

8. you quoted the answer you are looking for. I explicitly did those calculations for you. When I wrote <a,b,c> it is supposed to represent a column vector. The commas tell you when that component is done and to move to the next row.

Your function tells you to put in a vector, add the first two components together and keep the last one the same. it makes what was a 3 dimensional vector into a 2 dimensional one.

You bottom calculation is going in the right direction, you just add though man. Do you know how to add vectors? Just do it component wise. Add the first two terms together make that your new first term. Add the second terms together make that your new second term.

Look carefully at what I posted and write it down in proper column forms and see what I did, it is all there for you. Maybe you should practice doing this with just some actual numbers in there instead of x,y,z like 3,5,7 or something to see how the argument works. I dunno what else to tell you.

9. A clue that something is not right is that in your last vector you have just created 3 things out of nowhere. Where did you get $x_3, y_3, z_3$?

10. Originally Posted by Gamma
You are way fancier with the LaTeX code than me, I apologize in advance.
Let $u=$
Let $v=$
Let $a\in \mathbb{R}$
You must show that (1) $T(u + v)=T(u)+T(v)$ and (2) T(au)=aT(u).

(1)
$T(u+v)=T()=$
$T(u) + T(v) = + = $

(2)
$T(au)=T()= =a=aT(u)$

Am i on the right? This is what i did.
$T(u + v) = T(u) + T(v)$
$T(u + v) = T\left[ {\begin{array}{*{20}{c}}
{{x_1} + {x_2}} \\
{{y_1} + {y_2}} \\
{{z_1} + {z_2}} \\
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{x_1} + {y_1}} \\
{{z_1}} \\
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{x_2} + {y_2}} \\
{{z_2}} \\
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{x_3} + {y_3}} \\
{{z_3}} \\
\end{array}} \right]
$

$
T(u) + T(v) = \left[ {\begin{array}{*{20}{c}}
{{x_1} + {x_2}} \\
{{x_3}} \\
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{y_1} + {y_2}} \\
{{y_3}} \\
\end{array}} \right] = ?
$

I don't know what to do next or even if what i did is right? If i am wrong could you please tell me where i went wrong?
Thanks
KC

11. Good start, do you actually understand why that is correct? Or what you are even trying to prove?

You were almost there in showing equality between the two things, you just gotta add your two 2-vectors you got from T(u) +T(v) and I think you will see that you do indeed get the same thing. Remember the goal is to show T(u+v)=T(u)+T(v) AND T(au)=aT(u)

Once you understand that it should be a cakewalk showing T(au)=aT(u)

12. Originally Posted by Gamma
Good start, do you actually understand why that is correct? Or what you are even trying to prove?

You were almost there in showing equality between the two things, you just gotta add your two 2-vectors you got from T(u) +T(v) and I think you will see that you do indeed get the same thing. Remember the goal is to show T(u+v)=T(u)+T(v) AND T(au)=aT(u)

Once you understand that it should be a cakewalk showing T(au)=aT(u)
$
T(u) + T(v) = \left[ {\begin{array}{*{20}{c}}
{{x_1} + {y_1}} \\
{{z_1}} \\
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{x_2} + {y_2}} \\
{{z_2}} \\
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{x_1} + {x_2} + {y_1} + {y_2}} \\
{{z_1} + {z_2}} \\
\end{array}} \right]
$
Because they are equal it proves 1

$
T(cu) = T\left[ {\begin{array}{*{20}{c}}
{cx} \\
{cy} \\
{cz} \\
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{cx + cy} \\
{cz} \\
\end{array}} \right] = c\left[ {\begin{array}{*{20}{c}}
{x + y} \\
z \\
\end{array}} \right] = cT(u)
$

I really understand this one now. Thanks a lot.

13. BINGO, good to hear.

14. Originally Posted by igodspeed
Let
$\{T: R^n \to R^n |T{\text{ is linear\} }}$

$\left[ {\begin{array}{c}
x \\
y \\
z \\
\end{array}} \right] = \left[ {\begin{array}{c}
{x + y} \\
z \\
\end{array}} \right]

$
Thread closed due to OP's habit of deleting posts after getting replies.