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

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I'll be back in a sec. Brewing some coffee. Would you like some?
Yeah, I will in maybe two hours =P
Well you need to remember that a basis for any vectorial space generates it and it's vectors are linear independent.

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Other answers:

And 3x3 skew-symmetric matrix is something like this:
Hmm thinking...
Do you have any idea to begin with?
Mm you need three matrices am I right?
Well the dimension is 3 haha
I am stupid haha
and I need a longer explanation b/c A=-A^T isn't gonna cut it
soy estupido haha
uff I almost burn my brain hehe
I told you linear algebra's no good!
Well I think that the three vectors are \[\left[\begin{matrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{matrix}\right]\]
Hmmm you're right
Is the idea of a basis clear to you?
Like you know why (1,0,0), (0,1,0),(0,0,1) is the basis of R^3?
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.
like \[\vec{v}=a\hat{i} + b\hat{j} + c\hat{k}\]
well that is the idea in a graphic form.
If you want to obtain a basis for the 3x3 skew symmet
ric matrices you need to know that is the general form of matrix of that kind.
okay, I'm following what you're saying
the definition says that the elements of an skew symmetric matrix are such as: \[a_{ij}=-a_{ji}\]
It seems that that doesn't tell us a lot about those matrices, but you can figure out what you have on the diagonal.
if i = j, then \[a_{ii}=-a_{ji}\]
which is true if the diagonal is made up zeros.
and I think that's all. An 3x3 skew symmetrical matrix has this form:
Here's something...
\[\left[\begin{matrix}0 & a_{12} & a_{13} \\ -a_{12} & 0 & a_{23}\\ -a_{13} & -a_{23} & a_{33} \end{matrix}\right]\]
Looks like it's all zeros down the diagonal.
shame on my haha
yeah a_33 = 0
you're a bad math teacher haha! :-p
yeah! I'm terrible =(
but it's fun!
well now you can split that matrix on three parts:
Yep this part is where I need explaining...
\left[\begin{matrix}0 & a_{12} & 0 & \\ -a_{12} & 0 & 0\\ 0&0&0\end{matrix}\right]
\[\left[\begin{matrix}0 & 0 & a_{13} & \\ 0 & 0 & 0\\ -a_{13}&0&0\end{matrix}\right]\]
\[\left[\begin{matrix}0 & 0 & 0 & \\ 0 & 0 & a_{23}\\ 0&-a_{23}&0\end{matrix}\right]\]
Is it clear why I did that?
Oh I see. So each part of the basis will be a 3x3 matrix?
And there will be (3) 3x3 matrices
yeah! it has to bee a 3x3 matrix
I asked on cramster and some (puta) was really insulting to me AND she gave me the wrong answer.
what is cramster? and puta?
we are not finished yet.
you can do this:
I thought it meant b)it(ch but I guess it's mexican slang for prostitute or whore
but you get the idea.
\[\left[\begin{matrix}0& 1 & 0 \\ -1 & 0 &0\\ 0&0&0\end{matrix}\right]\]
ahh I thought I did haha.
\[\left[\begin{matrix}0& 0 & 1 \\ 0 & 0 &0\\ -1&0&0\end{matrix}\right]\]
and \left[\begin{matrix}0& 0 & 0 \\ 0 & 0 &1\\ 0&-1&0\end{matrix}\right]
You multiply by the scalars a,b,c each of those matrices and form a linear combination.
and the result is a skew symmetrical matrix
I'm sorry If my explanation was awful, and my english worst =P
So the dimension of all skew-symmetric matrices is 2?
or that's only the dimension of the basis?
the dimesion is the number of vectors of the basis. so it's 3.
hahaha I am so stupid!
WOW. I can't believe I asked that.
if you have an 4x4 skew symmetrical matrix the dimension of the basis is 4.
:-) yeah yeah yeah
Don't worry, don't judge yourself so severely.
Do you have any doubt regarding this brinethery?
I have another question for you! No I don't have any doubt. You did a great job explaining.
Linear Algebra?

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