# Is there a basis for the space of density matrices?

Let $$H_A$$ be a finite dimension Hilbert space. I consider matrices of this space, thus the space $$\mathcal{L}(H_A)$$.

I would like to know (I think I have read it somewhere but I'm not sure) if there exist a basis of this space composed of density matrices ?

I remind that density matrices are operator hermitic, semi definite positive, of trace $$1$$.

I think I can show that any matrix in $$\mathcal{L}(H_A)$$ can be written as a sum of density matrices doing the following:

First, any hermitic $$H$$ matrix is a sum of density matrices. Indeed, considering $$|\psi_i \rangle$$ an orthonormal basis in which $$H$$ is diagonal, we have, with $$\lambda_i \in \mathbb{R}$$:

$$H=\sum_i \lambda_i |\psi_i \rangle \langle \psi_i |$$

Then, any matrix $$A$$ can be written as:

$$A=H_1+i H_2$$

Where $$H_1$$ and $$H_2$$ are hermitic.

Then, $$A$$ can be written as a sum of density matrices, the coefficients being either real or pure imaginary.

Now, how to prove that there exist a basis of density matrices in which any operator $$A$$ can be decomposed ? If there is a simple example of such basis I would like to see it as well (decomposed in the canonical basis $$|i\rangle \langle j|$$).

What confuses me a little bit and that I have forgotten from linear algebra basics is that I see that any $$A$$ can be written as sum of density matrices. Does that necesseraly implies that there is a basis of density matrices or not necesserally ?

• $H_2$ could be indefinite – Nick Alger Jan 16 '20 at 19:21
• @NickAlger I don't see what you mean – StarBucK Jan 16 '20 at 19:24
• – metamorphy Jan 25 '20 at 6:52
• @StarBucK Did I answer your question? – Yly Jan 30 '20 at 23:31
• @Yly yes thank you very much. I will just have a question about your last paragraph in the few days, but you mainly answered it so I accept it ! Thank you very much. – StarBucK Jan 31 '20 at 0:09

• $$\left|j\rangle \langle j\right|$$
• $$\frac{1}{2}\left(\left|j\rangle+\left|k\rangle\right) \left( \langle j\right| + \langle k\right|\right)$$
• $$\frac{1}{2}\left(\left|j\rangle+i\left|k\rangle\right) \left( \langle j\right| - i\langle k\right|\right)$$
Where $$j, k$$ range over some fixed orthonormal basis set for $$H_A$$, with $$j< k$$, and where $$i$$ is the imaginary unit. These are clearly symmetric with unit trace, and they are positive semidefinite because they are all of the form $$\left|v\rangle \langle v\right|$$ for some $$v$$. To see that they form a basis for the set of operators on $$H_A$$, note that there are the right number of them ($$n^2$$, where $$n$$ is the dimension of $$H_A$$), so if we can show that they span the space of matrices then we are done.
The diagonal matrices are clearly attained using just the first type of matrices $$\left|j\rangle \langle j\right|$$. Similarly an off-diagonal matrix $$\left|j\rangle \langle k\right|$$ can be written as $$\frac{1}{2}\left(\left|j\rangle+\left|k\rangle\right) \left( \langle j\right| + \langle k\right|\right) + \frac{i}{2}\left(\left|j\rangle+i\left|k\rangle\right) \left( \langle j\right| - i\langle k\right|\right) - \left|j\rangle \langle j\right| - \left|k\rangle \langle k\right|$$, showing that the above set of matrices spans the space.
More generally, any time you have a subset $$S$$ of a vector space $$V$$ whose span contains the entire space, you can choose a basis from said subset $$S$$. The proof proceeds by induction. Choose an element $$b\in S$$ of the subset to be the first candidate basis element. Then, given any collection of candidate basis elements $$b_1, \dots, b_k$$, if the number of them doesn't equal the dimension $$n=\dim(V)$$ of the vector space, there must be some $$x\in S$$ which is linearly independent of $$b_1,\dots,b_k$$, because if not, then the span of $$S$$ would have strictly lower dimension than $$V$$! So we can keep adding elements to our candidate basis set $$b_1,\dots,b_k$$ until we get $$n$$ linearly independent elements, which are necessarily a basis for $$V$$.