Question about linear transformations

[bonildo]

Summary:: linear transformations

Hello everyone, firstly sorry about my English, I'm from Brazil.

Secondly I want to ask you some help in solving a question about linear transformations.

Here is the question:Consider the linear transformation described by the matrix $\mathsf{A} \in \Re ^{2x2} given by:$

$A = \begin{pmatrix} 1 & 1 \\ -1 & 1 \\ \end{pmatrix}$

a) Find the representation of the linear transformation in the basis V={v1,v2}, where v1=transpose(1,1) , v2=transpose(2,0)

My approach:

Choosing and arbitrary vector in the vector space that V span then it can be write as a linear combination of the basis:

$v=(x,y)=a1(1,1)+a2(2,0)$

Applying T on both sides:

$T(v)=T((x,y))=a1T(1,1)+a2T(2,0)$

Finding T(1,1) and T(2,0):

$T(1,1)=A*(1,1) =(2,0)$
$T(2,0)=A*(2,0) = (2,-2)$

then:

$T((x,y))= (2 a1 + 2 a2, -2 a2)$

Solving for a1 and a2:

$a1=(x+y)/2$
$a2=-y$

and finally T(x,y):
$T(x,y)=(x+y)/2 (2,0) +(-y)(2,-2) = (x-y,2y)$But when I substitute T(x,y) with (1,1) I don't get the same answer as A*(1,1) . Can someone help me with it ?

T(1,1)=(1-1,2*1) =(0,2)
and
A*(1,1) = (2,0)

[Moderator's note: Moved from a technical forum and thus no template.]

 

[BvU]

Hi,

I don't understand what you are doing. Could you clarify your symbols ?
e.g if you write $$T(1,1)=A*(1,1) =(2,0)$$ it looks to me as if you think $T = A$.

I also get another result:$$
A \begin{pmatrix} 1\\ 1\\ \end {pmatrix} =
\begin{pmatrix}
\phantom - 1 & 1 \\
-1 & 1 \\
\end{pmatrix}\begin{pmatrix} 1\\ 1\\ \end {pmatrix}=\begin{pmatrix} \phantom - 1+1\\ -1+1\\ \end {pmatrix} =\begin{pmatrix} 2\\ 0\\ \end {pmatrix}$$which in the basis V is equal to $v_2$. Similarly$$
A \begin{pmatrix} 2\\ 0\\ \end {pmatrix} = \begin{pmatrix} \phantom - 2\\ -2\\ \end {pmatrix} = 2 v_1 - 2v_2 $$So in the basis V, I would expect $$A'=
\begin{pmatrix}
0 &\phantom - 2 \\
1 & -2 \\
\end{pmatrix}$$

 

[bonildo]

Hi,

I don't understand what you are doing. Could you clarify your symbols ?
e.g if you write $$T(1,1)=A*(1,1) =(2,0)$$ it looks to me as if you think $T = A$.

I don't mean T=A , I mean T((1,1))=A*(1,1). In other words, the transformation applied to the vector (1,1) is equal to matrix A times (1,1). But I'm not sure if this equation is right... Did you get it ?

 

[PeroK]

I don't mean T=A , I mean T((1,1))=A*(1,1). In other words, the transformation applied to the vector (1,1) is equal to matrix A times (1,1). But I'm not sure if this equation is right... Did you get it ?

You need to be careful with notation when dealing with more than one basis. We have the first basis we are given:
$$e_1 \leftrightarrow \begin{pmatrix} 1\\ 0\\ \end {pmatrix}, e_2 \leftrightarrow \begin{pmatrix} 0\\ 1\\ \end {pmatrix}$$
In which the linear transformation $T$ is represented by the matrix:
$$T \leftrightarrow A = \begin{pmatrix}
1 & 1 \\
-1 & 1 \\
\end{pmatrix}
$$
Now, you have a second basis. I'll use the notation that vectors and linear transformations represented in this basis are indicated by a prime $'$. The basis vectors are:
$$v_1 \leftrightarrow \begin{pmatrix} 1\\ 0\\ \end {pmatrix}' \leftrightarrow \begin{pmatrix} 1\\ 1\\ \end {pmatrix}, v_2 \leftrightarrow\begin{pmatrix} 0\\ 1\\ \end {pmatrix}' \leftrightarrow \begin{pmatrix} 2\\ 0\\ \end {pmatrix}$$
And the linear transformation $T$ is represented by the matrix:
$$T \leftrightarrow A' = \begin{pmatrix}
a & b\\
c & d \\
\end{pmatrix}'
$$
Where you have to find $a, b, c, d$.

Unless you use the primed notation, you are going to get confused. Does that make sense?