Basis of a real hermitian matrix vector space with complex entries

[_Andreas]

Homework Statement

Let V be the \mathbb{R}-vector space \mbox{Herm}_n( \mathbb{C} ). Find \dim_{\mathbb{R}} V.

The Attempt at a Solution

I'd say the dimension is 2n(n-1)+n=2n^2-n, because all entries not on the main diagonal are complex, so you have n(n-1) entries which you have to split up in two (the scalars are real), and n real entries on the main diagonal (which you don't have to split up in two). However, the paper I have says that \dim_{\mathbb{R}} V is equal to n^2. I can't see how that could be correct. Have I misunderstood something?

 

[Dick]

The entries on opposite sides of the diagonal also have to be complex conjugates of each other. So you can really only choose one side freely. I'd count that as 2*(n^2-n)/2+n.

 

[Kreizhn]

You can simply think of taking the nxn Hermitian matrices, and stacking its rows on top of one another. This "map" will allow you to identify \mathrm{Herm}_n(\mathbb{C}) with \mathbb{C}^{n^2} since the matrix will have n² elements. All of the basic rules for the vector space hold since we're only considering scalar multiplication and addition, which is done component-wise (note that this map only becomes tricky if you're considering \mathrm{Herm}_n(\mathbb{C}) as an algebra, in which case you need to define your second binary operator in a special way). Thus you've reduced the question to, "what is the dimension of \mathbb{C}^{n^2} when viewed as a real vector space?" Well this is just trivially n². Though it seems that they've ignored restrictions on the definition of a Hermitian matrix.

 

[Kreizhn]

Yeah, including the restrictions for Hermiticity, I would only count

\frac{1}{2} n^2 + \frac{1}{2} n

Edit: I'm not too certain how you got 2n^2 - n since in the unrestricted case (and hence upper bound) case we have n², but

for n>1, 2n²-n > n²

 

[_Andreas]

Much appreciated help. Thanks to both of you!