Linear Algebra – Constructing Basis for Hermitian Matrices with Positive Semidefinite Ones

functional-analysislinear algebramatrices

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.

Best Answer

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.

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