# Thread: Linear transformation help

1. ## Linear transformation help

I have attached the question on the post, and am really struggling to get this right

MATH HELP .pdf

My teacher gave this solution but i dont get how T(1) = -1 and T(t) = 1?

I know how to get the standar matrix, i.e by applying T(e1) and T(e2) to the transformation and I get [0 1 -1 0]

Can someone please explain this solution to me, i am really unable to follow it

And can someone please explain how he gest L(1+t) = 1-t? and how L(1-t) = -1-t ?

Thank you.

2. ## Re: Linear transformation help

There is NO question in your attachment, just what looks like the solution to a problem. I presume that they know that T(1)= -t and T(t)= 1 because of the way T is defined in the problem itself. If you were to post that, we might be able to point out exactly where it gives the definition of T.

3. ## Re: Linear transformation help

The question is at the top, its link, but i'll attach here as well

4. ## Re: Linear transformation help

Oh, sorry, I completely missed that! It says that T(a+ bt)= b- at. T(1)= T(1+ 0t)= 0- 1t= -t since a= 1 and b= 0. T(t)= T(0+ 1t)= 1- 0t= 1 since a= 0 and b= 1.

To find a matrix representation of a linear transformation in a given basis, apply the linear transformation to each basis vector in turn, the write the results as a linear combination of the basis vectors. The coefficients of the linear combinations give the columns of the matrix.

Here, the first basis is "{1, t}". As we saw before, T(1)= -t= 0(1)+ (-1)t and T(t)= 1(1)+ 0(t). The matrix representation in this basis is $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$. In this basis we would represent the general "a+ bt" as the vector $\begin{bmatrix}a \\ b\end{bmatrix}$ and you can see that $\begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}= \begin{bmatrix}b \\ -a\end{bmatrix}$ which represents b- at.

Do the same for the basis {1+ t, 1- t}. T(1+ t)= 1+ t, since a= b= 1 and T(1- t)= -1+ t since a= -1 and b= 1. Of course, 1+ t= 1(1+ t)+ 0(1- t) and -1+ t= 0(1+t)- 1(1- t) so the matrix representation in this basis is $\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$.