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.


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.

  • $\begingroup$ 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? $\endgroup$
    – joriki
    May 28, 2012 at 7:05
  • $\begingroup$ you are right. i wanted an explicit method for constructing such basis. otherwise i am more or less sure there are such basis $\endgroup$
    – RSG
    May 28, 2012 at 12:40
  • $\begingroup$ "the (real) linear combination of positive semi-definite matrices spans the space of hermitian matrices". How do you get $\begin{pmatrix}0 & i \\ -i & 0\end{pmatrix}$? $\endgroup$
    – user31373
    May 28, 2012 at 14:42
  • $\begingroup$ $\frac{1}{2}\begin{pmatrix} 1 & i\\-i & 1\end{pmatrix}-\frac{1}{2}\begin{pmatrix} 1 & -i\\i & 1\end{pmatrix}$ $\endgroup$
    – RSG
    May 28, 2012 at 16:04
  • 2
    $\begingroup$ 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. $\endgroup$
    – user31373
    May 28, 2012 at 19:55

1 Answer 1


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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