MHB Dance's question at Yahoo Answers (T(A)=A-A^T)

[Fernando Revilla]

Here is the question:

T(A) = A - A^T in R^2
using the basis
[1 0; 0 1], [0 0; 0 1], [0 1; 1 0], [0 -1; 1 0]

Here is a link to the question:

How to find the coordinate matrix of this transformation? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Dance,

Denote $B=\{e_1,e_2,e_3,e_4\}$ where $$e_1=\begin{bmatrix}{1}&{0}\\{0}&{1}\end{bmatrix}, \;e_2=\begin{bmatrix}{0}&{0}\\{0}&{1}\end{bmatrix}, \;e_3=\begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}, \;e_4=\begin{bmatrix}{0}&{-1}\\{1}&{0}\end{bmatrix}$$

Then, $$T(e_1)=e_1-e_1^T=\begin{bmatrix}{1}&{0}\\{0}&{1}\end{bmatrix}-\begin{bmatrix}{1}&{0}\\{0}&{1}\end{bmatrix}= \begin{bmatrix}{0}&{0}\\{0}&{0}\end{bmatrix}=0e_1+0e_2+0e_3+0e_4\\T(e_2)=e_2-e_2^T=\begin{bmatrix}{0}&{0}\\{0}&{1}\end{bmatrix}-\begin{bmatrix}{0}&{0}\\{0}&{1}\end{bmatrix}= \begin{bmatrix}{0}&{0}\\{0}&{0}\end{bmatrix}=0e_1+0e_2+0e_3+0e_4\\T(e_3)=e_3-e_3^T=\begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}-\begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}= \begin{bmatrix}{0}&{0}\\{0}&{0}\end{bmatrix}=0e_1+0e_2+0e_3+0e_4\\T(e_4)=e_4-e_4^T=\begin{bmatrix}{0}&{-1}\\{0}&{1}\end{bmatrix}-\begin{bmatrix}{0}&{1}\\{-1}&{0}\end{bmatrix}= \begin{bmatrix}{0}&{-2}\\{2}&{0}\end{bmatrix}=0e_1+0e_2+0e_3+2e_4$$ Transposing coefficientes: $$A=[T]_B=\begin{bmatrix}{0}&{0}&{0}& 0\\{0}&{0}&{0}& 0\\{0}&{0}&{0}& 0\\{0}&{0}&{0}& 2\end{bmatrix}$$ If you have further questions, you can post them in the Linear and Abstract Algebra section.