# Do they form a vector space?

• Oct 16th 2010, 03:15 AM
Do they form a vector space?
Do n-by-n traceless matrices form a vector space over the field C of complex numbers? Why or why not? If yes, show a basis and give the dimension of the vector space!

Over the field R of real numbers, I know that they form a vector space:
Tr(A)+Tr(B)=Tr(A)+Tr(B)
TR(c*A)=c*Tr(A)
0 matrix is in the vector space
-A matrix is in
But I don't know what is a basis over real numbers.
And I don't even know if they form a vector space over the field C of complex numbers. I suppose they do, but I am not sure, I can't prove it.

• Oct 16th 2010, 05:43 AM
HallsofIvy
The only difference between "over the field R" and "over the field C" is that, in the second, the scalar, "c", you multiply by could be a complex number. Is TR(c*A)= c*Tr(A) when c is any complex number?
• Oct 16th 2010, 07:02 AM
Quote:

Originally Posted by HallsofIvy
The only difference between "over the field R" and "over the field C" is that, in the second, the scalar, "c", you multiply by could be a complex number. Is TR(c*A)= c*Tr(A) when c is any complex number?

Yes, I think it is true that TR(c*A)= c*Tr(A) when c is any complex number. Am I right? And if so, does it mean that the basis and the dimension of the vectorspace are the same over the field R and over the field C?
Thanks for helping me!
• Oct 17th 2010, 09:46 AM