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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.