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let us consider $n\times n$ hermitian matrices. They form a real space. Now we know that any such matrix $A$ can be written as $A=A_+-A_-$, where $A_\pm$ are positive semidefinite matrices. Thus we can say that the (real) linear combination of positive semi-definite matrices spans the space of hermitian matrices. My question is, can we construct any basis for the space of hermitian matrices such that each basis element is a positive semidefinite matrix. please help or refer some literature for it.

ADDED:

seeing the comment of Joriki I have decided to add a few more lines regarding my earlier (failed) approach, in the hope that someone can help me (in completing the line of argument, if possible; or by finding a fault in my argument). I can diagonalise and separate the positive and negative part. now let $A=UDU^*$, where $U$ is an unitary operator and $D$ is the diagonal matrix. This again can be written as $A=UD_1U^*-UD_2U^*$ where $D_i$ are diagonal matrices with all entries $\geq0$. hence, if we take a such a positive diagonal matrices and only consider unitary group action on it, we are going to find all the hermitian operators. in particular, i tried to take diagonal matrix $D_j$ ($j$-th entry $1$, others $0$) and applied unitary group. this method seemed to fail here, as i could not get any meaningful basis out of these actions.

glS
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RSG
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  • I'm not sure I understand the question correctly. You've already stated that the positive semi-definite matrices span the hermitian matrices. It's always possible to select a basis from a spanning set. Were you unaware of this fact; or is your question whether we can explicitly construct a specific such basis? – joriki May 28 '12 at 07:05
  • you are right. i wanted an explicit method for constructing such basis. otherwise i am more or less sure there are such basis – RSG May 28 '12 at 12:40
  • "the (real) linear combination of positive semi-definite matrices spans the space of hermitian matrices". How do you get $\begin{pmatrix}0 & i \ -i & 0\end{pmatrix}$? –  May 28 '12 at 14:42
  • $\frac{1}{2}\begin{pmatrix} 1 & i\-i & 1\end{pmatrix}-\frac{1}{2}\begin{pmatrix} 1 & -i\i & 1\end{pmatrix}$ – RSG May 28 '12 at 16:04
  • @Leonid: More generally, any hermitian matrix can be diagonalized with its real eigenvalues on the diagonal, and then $A_+$ is obtained from a diagonal matrix containing the positive eigenvalues and $A_-$ is obtained from a diagonal matrix containing the absolute values of the negative eigenvalues. – joriki May 28 '12 at 19:29
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    Thanks, I was confused by something else. Returning to the original question: we can begin with $\begin{pmatrix}1&0\0&0\end{pmatrix}$, $\begin{pmatrix}0&0\0&1\end{pmatrix}$, $\begin{pmatrix}0&1\1&0\end{pmatrix}$, $\begin{pmatrix}0&i\-i&0\end{pmatrix}$ and add the identity matrix to the non-diagonal matrices, thus making them positive definite. Works in higher dimensions too. –  May 28 '12 at 19:55
  • @LeonidKovalev it certainly works (with probably a slight modification), if i have understood your comment correctly. we can consider any hermitain operator $A$ (need not be positive) and consider $ cI+A$ where $c\geq0$ and $I$ is the identity matrix. then there will always be a $c$ such that the combination is positive semidefinite. in your case, it is simpler. consider generalised non-diagonal Pauli matrices and your method will work. for the diagonal ones, use the above method. (here i have assumed the generalisation to be of Gell-Mann matrices type) – RSG May 29 '12 at 04:56
  • @Leonid or rsg, you could write up Leonid's comment as an answer so the question doesn't remain unanswered. – joriki May 29 '12 at 07:22

1 Answers1

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The space of hermitian $n\times n$ matrices is spanned by the $n$ matrices with a single $1$ on the diagonal, the $n(n-1)/2$ matrices with a single pair of $1$s at corresponding off-diagonal elements and the $n(n-1)/2$ matrices with a single pair of $\mathrm i$ and $-\mathrm i$ at corresponding off-diagonal elements. The diagonal matrices are positive semidefinite, and the remaining matrices can be made positive semidefinite by adding $1$ to the two diagonal elements corresponding to the non-zero off-diagonal elements. Since adding one element of a linearly independent set to another doesn't render the set linearly dependent, the result is again a basis of the space of hermitian matrices.

joriki
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