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OpenStudy (anonymous):

Find a basis for the space V of all skew-symmetric 3x3 matrices. What is the dimension of V?

OpenStudy (anonymous):

I'll be back in a sec. Brewing some coffee. Would you like some?

OpenStudy (anonymous):

Yeah, I will in maybe two hours =P

OpenStudy (anonymous):

Well you need to remember that a basis for any vectorial space generates it and it's vectors are linear independent.

OpenStudy (anonymous):

And 3x3 skew-symmetric matrix is something like this:

OpenStudy (anonymous):

\[A=-A^T\]

OpenStudy (anonymous):

:-)

OpenStudy (anonymous):

Hmm thinking...

OpenStudy (anonymous):

Do you have any idea to begin with?

OpenStudy (anonymous):

Mm you need three matrices am I right?

OpenStudy (anonymous):

Well the dimension is 3 haha

OpenStudy (anonymous):

I am stupid haha

OpenStudy (anonymous):

and I need a longer explanation b/c A=-A^T isn't gonna cut it

OpenStudy (anonymous):

soy estupido haha

OpenStudy (anonymous):

uff I almost burn my brain hehe

OpenStudy (anonymous):

I told you linear algebra's no good!

OpenStudy (anonymous):

Well I think that the three vectors are \[\left[\begin{matrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{matrix}\right]\]

OpenStudy (anonymous):

Hmmm you're right

OpenStudy (anonymous):

Is the idea of a basis clear to you?

OpenStudy (anonymous):

Like you know why (1,0,0), (0,1,0),(0,0,1) is the basis of R^3?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

Well then you know that the basic idea behind a basis is that you can get any vector of the space by forming a linear combination.

OpenStudy (anonymous):

like \[\vec{v}=a\hat{i} + b\hat{j} + c\hat{k}\]

OpenStudy (anonymous):

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