Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 7:05
    
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 $\begin{pmatrix}0 & i \\ -i & 0\end{pmatrix}$? –  user31373 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
1  
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. –  user31373 May 28 '12 at 19:55
show 3 more comments

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.