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Linear Transformations and their matrices 
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#1
Mar2409, 08:13 PM

P: 33

The problem is
T(x + yi) = x  yi Show that this is a linear transformation and find the matrix of the transformation using the following basis (1+i, 1i) ARGH I am having trouble with the complex numbers for some reason! To show that it is linear I have to show T(x + yi + a + bi) = x + (yi) + a + (bi) = x + a + ( i(y +b)) = T(x + yi + a +bi) = x + (yi) + a + (bi) T(k(x + yi)) = k(x + (yi)) = kx + k(yi) = T(k(x + (yi)) = kT(x + (yi)) Is this correct? And as for finding the matrix. OY! I know B = [ [T(1+i)] [T(1i)] ] So I know how to do it in theory, kind of I guess. But I just don't know how to start :/ 


#2
Mar2509, 01:19 AM

Mentor
P: 21,249

Start with the definition for a linear transformation. If T is a linear transformation from C to C, what conditions have to be satisfied?



#3
Mar2509, 02:02 AM

P: 33

Well the definition of a transformation is it's closed under addition and scalar multiplication, but I don't see how that helps. Is this regarding finding the matrix of the transformation?



#4
Mar2509, 06:55 AM

Math
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Thanks
PF Gold
P: 39,488

Linear Transformations and their matrices
Apply the linear transformation to each basis vector in turn. Write the result in terms of the basis. The coefficients are form the columns of the matrix. That's exactly what your "B= [[T(1+i)] [T(1i)]]" means. For example, here your given basis is 1+ i, 1 i. T(1+ i)= 1 i= 0(1+ i)+ 1(1 i). The first column of the matrix is [tex]\begin{bmatrix}0 \\ 1\end{bmatrix}[/tex]. What is the second column? 


#5
Mar2509, 08:38 AM

Mentor
P: 21,249

T(u + v) = T(u) + T(v) T(au) = aT(u) where for your problem, u and v are complex numbers, and a is a real number. 


#6
Mar2509, 01:26 PM

P: 33

To HallsofIvy,
So you're saying we take each basis, perform the linear transformation on the basis, and then we find a linear combination of the original basis that satisfies the transformation and that gives me my column? So the second column would be something like this T(1i) = (1 + i) = c1(1+i) + c2(1i) = 1(1+i) + 0(1i). so the second column would be 1 0 ? 


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