Basis of a real hermitian matrix vector space with complex entries

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  • #1
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Homework Statement



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

The Attempt at a Solution



I'd say the dimension is [tex]2n(n-1)+n=2n^2-n[/tex], because all entries not on the main diagonal are complex, so you have [tex]n(n-1)[/tex] 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 [tex]\dim_{\mathbb{R}} V[/tex] is equal to [tex]n^2[/tex]. I can't see how that could be correct. Have I misunderstood something?
 
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Answers and Replies

  • #2
Dick
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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
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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 [itex] \mathrm{Herm}_n(\mathbb{C}) [/itex] with [itex] \mathbb{C}^{n^2} [/itex] 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 [itex] \mathrm{Herm}_n(\mathbb{C}) [/itex] 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 [itex] \mathbb{C}^{n^2} [/itex] 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
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Yeah, including the restrictions for Hermiticity, I would only count

[tex] \frac{1}{2} n^2 + \frac{1}{2} n [/tex]

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

for n>1, 2n²-n > n²
 
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  • #5
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Much appreciated help. Thanks to both of you!
 

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