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!