Properties of linear transformation, did my professor make an error?

mr_coffee
Messages
1,613
Reaction score
1
Hello everyone, I'm studying an example my professor did, and it isn't making sense to me... here is the orignal matrix:
THe oringal matrix is:
T = [3s-t]
[t]...[2t+7s]
he wants to determine if the following trnasformation is Linear.
Here is what he wrote on the board:
http://img205.imageshack.us/img205/8454/lastscan0mm.jpg

Why does he say:
T[x*s1]
..[x*s2]

then when he finally proves that is passes the 2nd test of linear transformations, he writes:

x*T[s1]
...[t1]

when right up there he has s2 on the bottom, did he to say:
T[x*s1]
..[x*t1]
 
Last edited by a moderator:
Physics news on Phys.org
That's just a typo ofcourse. The whole subscript thing is also redundant, that's probably why he confused s2 and t1. s2 and t1 are both referred to as the second component of the vector.
 
So is it suppose to be:
T[x*s1]
..[x*t1]
?
I don't want to miss it on the exam, because this was the pratice exam
 
In the example you attached, replace each occurence of t1 with s2.

The result would then be,

T<br /> \begin{array}{|c|}<br /> \alpha s_1 \\<br /> \alpha s_2 \\<br /> \end{array}<br /> =<br /> \alpha T<br /> \begin{array}{|c|}<br /> s_1 \\<br /> s_2 \\<br /> \end{array}<br />
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top