Basis of positive matrices

[Zareon]

Consider an n-dimensional complex vector space.

The corresponding vector space of nxn matrices is n^2 dimensional.

I want to find out if there exists a basis for this space consisting of semipositive matrices.

 

The question seems simple, but I can't find a proof.

any help is appreciated.

[Dave]

Could you give the definition of semi-positive? I've not seen that term.

[D H]

Positive definite = all eigenvalues are real and positive.

Positive semidefinite = all eigenvalues are real and nonnegative.

[ajb]

By positive definite you exclude the possibility of zero as an eigenvalue, as for positive semidefinite you include zero as an eigenvalue. Is that right DH?

[D H]

Consider an n-dimensional complex vector space.

The corresponding vector space of nxn matrices is n^2 dimensional.

How exactly are you proposing to construct vectors out of matrices?

I want to find out if there exists a basis for this space consisting of semipositive matrices.

Start with the simplest case, 2x2 matrices.

 

By positive definite you exclude the possibility of zero as an eigenvalue, as for positive semidefinite you include zero as an eigenvalue. Is that right DH?

You also exclude complex numbers as an eigenvalue, which excludes most non-symmetric matrices.

[Country Boy]

How exactly are you proposing to construct vectors out of matrices?

?? The set of nxn matrices, indeed, even of general n by m matrices, forms a vector space with the usual addition of matrices and multiplication by numbers.

 

Start with the simplest case, 2x2 matrices.

 

 

You also exclude complex numbers as an eigenvalue, which excludes most non-symmetric matrices.

[alext87]

For every n-dimensional vector space there does exist a basis consisting of semi-definite matrices.

Proof: Take the canonical basis of the vector space. This is where we have matrices {E11,E12,...,E1n,E21,E22,...,E2n,...,...,En1,En2,...,Enn} where Eij is a matrices with all zero elements except the (i,j) entry.

All these matrices are semi-definite.

As the characteristic polynomial is x^n for all Eij where i is not equal to j and the characteristic poly is x^(n-1) * (x-1) for Eii 1<=i<=n.

Now all eigenvalues of a matrix are the roots of the characteristic poly i.e (in this case) just 0 and 1. So the matrices are all semi-definite.

 

Hope this helps.

 

AT