# Basis of a real hermitian matrix vector space with complex entries

1. Oct 10, 2008

### _Andreas

1. The problem statement, all variables and given/known data

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

3. 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?

Last edited: Oct 10, 2008
2. Oct 10, 2008

### 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.

3. Oct 10, 2008

### 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.

4. Oct 10, 2008

### 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

Last edited: Oct 10, 2008
5. Oct 11, 2008

### _Andreas

Much appreciated help. Thanks to both of you!