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An inner product space is a Hilbert space if and only if every Cauchy sequence converges in the vector space itself.That is, every complete inner product space is hilbert.

The inner product in this space is defined as: $(A,B)=Trace(A^\dagger B)$,where A,B are 2x2 traceless Hermitian matrices.

In order to prove that the above vector space is a Hilbert space, we can consider a Cauchy sequence and show that it converges into the space. But can we prove it in a more general way?

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  • $\begingroup$ Is it isomorphic to a simpler space? $\endgroup$
    – andypea
    Oct 31, 2016 at 10:09
  • $\begingroup$ Math SE would be the better for this question. $\endgroup$
    – Mass
    Oct 31, 2016 at 10:11
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    $\begingroup$ The question is meaningless if you do not define a symmetric real scalar product on that space. An important point is that we are speaking about a real Hilbert space, since a complex combination of Hermitian matrices is not Hermitian. $\endgroup$ Oct 31, 2016 at 10:59
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    $\begingroup$ "A vector space is a hilbert space iff every cauchy sequence converges in the vector space itself." This definition is not complete: "A vector space equipped with a scalar product is a Hilbert space iff every Cauchy sequence - defined with respect to the norm associated to the scalar product - converges in the vector space itself." $\endgroup$ Oct 31, 2016 at 11:01

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Let $H(n,\mathbb C) \subset M(n,\mathbb C)$ denote the vector space over $\mathbb R$ of Hermitian traceless $n \times n$ matrices. It is well-defined: As the trace is linear, a linear combination of traceless matrices is traceless too, and a real combination of Hermitian matrices is Hermitian as well.

Define $$\langle A| B \rangle := Re\: tr(A^\dagger B) \:, \qquad A,B \in H(n,\mathbb C) \:.$$ It is easy to prove that $\langle \cdot |\cdot \rangle : H(n,\mathbb C) \times H(n,\mathbb C) \to \mathbb R$ is $\mathbb R$-bilinear and symmetric from the elementary properties of the trace. It is also strictly positive, indeed $$\langle A|A \rangle = Re \: tr(A^\dagger A) = Re \left(\sum_{a,b=1}^n \overline{A_{ab}}A_{ab} \right) = \sum_{a,b=1}^n|A_{ab}|^2 \geq 0$$ where, evidently $\langle A|A \rangle=0$ implies $A_{ab}=0$ for all $a,b$.

All that proves that $\langle \cdot |\cdot \rangle$ is a well defined real scalar product over the real vector space $H(n,\mathbb C)$.

Proposition. $H(n,\mathbb C)$ equipped with the real scalar product $\langle \cdot |\cdot \rangle$ is a real Hilbert space.

Proof. $$||A||:= \sqrt{\langle A|A \rangle} = \sqrt{\sum_{a,b=1}^n|A_{ab}|^2}$$ is nothing but the standard Euclidean norm of $\mathbb C^{k}$ with $k=n^2$ which, viewed as a metric space with distance function $d(A,B) := ||A-B||$, is complete as is well known. In other words $M(n, \mathbb C) \equiv C^{n^2}$ is a complete metric space. To conclude it is enough to establish that $H(n,\mathbb C)$ is a closed subset of $M(n, \mathbb C)$ with respect to the topology induced by that distance.

In fact, a Cauchy sequence in $H(n, \mathbb C)$ is also Cauchy in $M(n, \mathbb C)$ which is complete and thus admits a limit therein. Assuming that $H(n, \mathbb C)$ is closed, it includes that limit, too. $H(n, \mathbb C)$ is therefore complete with respect to the norm induced from $\langle \cdot|\cdot \rangle$ and thus is a real Hilbert space as wanted.

Let us prove that $H(n, \mathbb C)$ is closed. The conditions $$\mbox{$tr A=0$ and $||A-A^\dagger|| =0$ for $A\in M(n, \mathbb C)$}$$ which define $H(n, \mathbb C)$, can be seen as defining the pre-images $f^{-1}(\{0\})$ and $g^{-1}(\{0\})$ of a pair of continuous maps $$f : M(n, \mathbb C) \ni A \mapsto tr A \in \mathbb C$$ and $$g: M(n, \mathbb C) \ni A \mapsto ||A-A^\dagger|| \in \mathbb R\:.$$ Continuity can be proved by direct inspection decomposing these functions into elementary continuous functions over $\mathbb R^p$. The two pre-images $f^{-1}(\{0\})$ and $g^{-1}(\{0\})$ are closed sets because they are pre-images of the closed set $\{0\}$ and the functions are continuous. The intersection $$H(n, \mathbb C) = f^{-1}(\{0\})\cap g^{-1}(\{0\}) $$ is closed because the intersection of closed sets is closed too. QED

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The $\textrm{trace}$ function is continuous from $\mathcal{M}_2(\mathbb{C})$ to $\mathbb{C}$ and $\{0\}$ is closed in $\mathbb{C}$, hence $\textrm{trace}^{-1}(0)$ is closed. A closed subspace of a Hilbert space is a Hilbert space, hence the space of traceless matrices is a Hilbert space. The subset of Hermitian matrices is not a subspace, but nevertheless is closed as being the reciprocal image of the null matrix of another continuous function from $\mathcal{M}_2(\mathbb{C})$ into itself: $f(M) = M^*-M$. Your set is the intersection of two closed sets, hence is closed too.

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