A question in prooving liniar operator

[transgalactic]

http://img228.imageshack.us/my.php?image=52788065ff7.jpg

the formula which i was given is fro a polinomial
although the operator is defined for normal 2X2 vector

i don't know how to approach to this formula?

 

[HallsofIvy]

I'm not sure what you mean by "for a polynomial". The formula is T(A)= A^T- A.

Perhaps you are thinking that A^T is a power? Even if it were, that would still be alright- if you can multiply matrices you can certainly take a matrix to a power- and a product of matrices still represents a linear transformation.

However, A^T is the standard notation for the 'transpose' of a matrix: basically you swap rows and columns. For a 2 by 2 matrix
$$\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]^T= \left[\begin{array}{cc} a & c \\ b & d\end{array}\right]$$
so that
[tex]\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]^T- \left[\begin{array}{cc}a & b \\ c & d\end{array}\right]= \left[\begin{array}{c c} a & c \\ b & d\end{array}\right]- \left[\begin{array}{cc}a & b \\ c & d\end{array}\right][/itex]<br /> $$= \left[\begin{array}{cc} 0 & c- b \\ b- c & 0\end{array}\right]$$[/tex]

 

[transgalactic]

no this sign is a little line like a derivative sign
no way it can't be a T
??

 

[HallsofIvy]

then I have no idea what it means- it's not a standard notation. Does your book give a definition? If not, ask your teacher.

 

[transgalactic]

i trust your judgement probably its a print mistake
and they ment T

 

[HallsofIvy]

It's my vision I'm not sure you should trust! Is this matrix over the real numbers or complex numbers? Sometimes a little "sword" superscript is use to represent the "Hermitian conjugate" where you take the transpose (switch rows to columns) and take the complex conjugate of all entries. Of course, if your matrix has only real number entries, that is the same as the transpose.