# How to find out the dimension of a given vector space?

What will be the dimension of a vector space $V =\{ a_{ij}\in \mathbb{C_{n\times n}} : a_{ij}=-a_{ji} \}$ over field $\mathbb{R}$ and over field $\mathbb{C}$?

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oh sorry for mistake – srijan May 8 '12 at 13:14
What have you tried? I would suggest you first try to solve the problem for $2\times 2$ or $3\times 3$-matrices. – Martin Wanvik May 8 '12 at 13:16
I was counting the number of independent entries. But got confusion. – srijan May 8 '12 at 13:19
Good idea. But surely, for $2\times 2$-matrices, there can be no room for confusion? Let me try a slightly different question: can you write down the general form of a matrix in $V$ for $n = 2$ and $n = 3$? – Martin Wanvik May 8 '12 at 13:23
Yes, in the sense that $a_{ij} = -a_{ji}$ - this seems surprisingly difficult to express in words, but here is an attempt: an off-diagonal element has to be equal in magnitude but of opposite sign as the element whose location is obtained by reflecting about the diagonal. – Martin Wanvik May 8 '12 at 13:32

What have you tried so far? The solution is pretty straight forward. You try to compose a basis consisting of matrices which are as simple as possible. Here "as simple as possible" usually means very few entries with values $\pm 1$ and zeros everywhere else.

What happens when you set $a_{ij}=1$ for $i\neq j$? You will also have $a_{ji}=-1$. What happens on the diagonal? Right, since $a_{ii}=-a_{ii}$ these entries must be zero. So a good candiadate for a basis (over $\mathbb C$) are those matrices with the property $a_{ij}=-a_{ji}=1$ for some $i<j$ and zeros everywhere else. Are they linear independent? Do they span your space?

If you want a basis over $\mathbb R$, you could take the same matrices as above and also those which have $\pm i$ in the places where $\pm 1$ used to be.

The dimension should be the number of entries below the diagonal (times 2 for $\mathbb R$).

Edit: A totally different approach is to consider the problem as follows (I'll treat the case that the field is $\mathbb C$): You have $n^2$ entries in a given matrix. Or in other words $n^2$ independent variables. Now your constraints can be viewed as linear equations in these variables. How many linear independent equations do we have: one for each pair $i<j$, which gives you $\binom n2$ equations and one for each $i=i$. In total we have $n^2$ variables with $\binom n2+n$ independent linear equations. Leaving you with $n^2-(\binom n2+n)$ degrees of freedom. The two results should obviously coincide.

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Without restrictions, the dimension of $V$ over $\mathbb{R}$ is $2n^2$. The 2 appears because the field of scalars is $\mathbb{R}$ and $\mathbb{C}$ as a two dimensional vector space over $\mathbb{R}$. The $n^2$ can be explained thinking on the elements of $V$ as complex matrices. The number of entries of a $n\times n$ matrix is $n^2$.

Your condition says that, thinking in terms of matrices, the diagonal elements are all zero, and knowing all the elements above or below the diagonal is sufficient to know all the elements of the matrix.

Now the tricky argument is the following: note that in the first row there are zero independent elements, in the second one, and in the nth $n-1$. So the number of independent elements, and therefore the dimension of the space is

$$\sum_{i=1}^{n-1} i= \frac{n(n-1)}{2}$$

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