It is true that the vector space of $n\times n$ Hermitian matrices is an $n^2-$dimensional real vector space and that one can find a basis for this space consisting exclusively of positive semi-definite matrices (see basis for hermitian matrices). My question is, if we have the linear combination $$ M=\sum_{k=1}^{n^2}x_kB_k $$ where $B_k$ is the aforementioned positive semi definite basis matrix, what are the restrictions on $x_k$ so that $M$ remains positive semi-definite? A sufficient condition is (I think) that all $x_k\geq 0$, but I don't think it is necessary. If $B_k$ were real numbers, the solution to that problem would be one of the two halfspaces into which $\mathbb{R}^{n^2}$ is divided into by the hyperplane $\sum_{k=1}^{n^2}x_kB_k=0$. Could there be an analogy with this case, when $B_k$ belong to the vector space of Hermitian matrices?

  • $\begingroup$ I think it is possible to choose $B_k$ such that the condition $x_k \geq 0$ is necessary. However, in general, the exact conditions will depend on the choice of basis. $\endgroup$ – Omnomnomnom May 13 '15 at 13:02
  • $\begingroup$ @AlgebraicPavel is it possible to select such a basis? $\endgroup$ – Omnomnomnom May 13 '15 at 14:17
  • $\begingroup$ @AlgebraicPavel Are you sure about the dimension? One has to select $n$ real diagonal elements, and $n(n-1)/2$ complex elements, which adds up to $n^2$ real parameters. BTW, I am using as a basis the construction of math.stackexchange.com/questions/150643/… , therefore the condition $B_iB_j=0$ is not fulfilled automatically. $\endgroup$ – Bryson of Heraclea May 13 '15 at 14:22
  • $\begingroup$ @Omnomnomnom Good point. It does not exist unless $n=1$ (the trivial case). There are at most $n$ nonzero HPSD matrices with this property and, well, $n<n^2$ otherwise :-) $\endgroup$ – Algebraic Pavel May 13 '15 at 14:55
  • $\begingroup$ @BrysonofHeraclea No I'm not, that's why I removed the comment. $\endgroup$ – Algebraic Pavel May 13 '15 at 14:55

Thm: Let $B_i\in M_n$ be a positive semidefinite Hermitian matrix for $1\leq i\leq s$. If $\sum_{i=1}^sa_iB_i$ is positive semidefinite iff $a_i\geq 0$ then $s\leq n$.

Proof: If $\Im(B_2)\subset\Im(B_1)$ then there is a small $a_2>0$ such that $B_1-a_2B_2$ is positive semidefinite. Thus, $\Im(B_2)$ is not a a subset of $\Im(B_1)$ and $rank(B_1+B_2)>rank(B_1)$. If $\Im(B_3)\subset\Im(B_1+B_2)$, then there is a small $a_3>0$ such that $B_1+B_2-a_3B_3$ is positive semidefinite. Thus, $\Im(B_3)$ is not a a subset of $\Im(B_1+B_2)$and $rank(B_1+B_2+B_3)>rank(B_1+B_2)$. We can repeat the argument $s$ times and if $s>n$ then $rank(B_1+\ldots+B_{n+1})>n$, which is impossible. So $s\leq n$.

  • $\begingroup$ Thank you very much for the reply! This is a significant step towards the more general answer. By $\mathfrak{I}(B_i)$ do you mean the range (columnspace) of the matrix $B_i$ ? $\endgroup$ – Bryson of Heraclea May 18 '15 at 7:27
  • $\begingroup$ Hi @BrysonofHeraclea. Yes you are right, $\Im(B_i)$ is the column space. $\endgroup$ – Daniel May 18 '15 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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