A linear combination $ax + by$ is called conic(al) if $a, b \ge 0$ (cf. section 2.1.5 of Boyd, Vandenberghe). I.e. conic(al) combinations are just linear combinations where the coefficients are restricted to be non-negative.

Here I am treating the space of $2 \times 2$ Hermitian (complex) PSD matrices as a real vector space (i.e. scalars are real numbers, so saying coefficients are $\ge 0$ makes sense).

Question: Can every Hermitian PSD (positive semi-definite) matrix be written as the conic combination of rank-1 Hermitian PSD matrices? If so, is there a reference or a simple argument?

Attempt: Under that assumption, and via trial and error, I think the following might be a minimal spanning set for the $2 \times 2$ complex Hermitian PSD cone. It seems like none of them can be written as conic combinations of each other (even though they're linearly dependent), so all that would need to be shown is that every PSD matrix can be written as a conic combination of these six matrices.

$$ (A) \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

$$ (B) \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$

$$ (C) \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

$$ (D) \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

$$ (E) \begin{bmatrix} 1 & i \\ -i & 1 \end{bmatrix}$$

$$ (F) \begin{bmatrix} 1 & -i \\ i & 1 \end{bmatrix}$$

Note that a linear basis for Hermitian matrices should have $n^2 = 4$ elements, whereas the above has $2n^2 - n$ elements. This is reflected in how the above set is linearly dependent, e.g. (C) = -(D) + 2(A) + 2(B), and (E) = -(F) + 2(A) + 2(B). But clearly neither of these are conic(al) combinations.

Obviously being linearly independent implies also being conically independent. So showing that the above set is minimal (assuming that every PSD matrix can be writen as a conical combination of them) requires showing that (D) and (F) cannot be written as conic(al) combinations of (A), (B), (C), (E), and/or that (C) and (E) cannot be written as conic(al) combinations of (A), (B), (D), (F).

This might follow from how (A), (B), (C), (E) form a basis for Hermitian matrices, as do (A), (B), (D), (F), and so by uniqueness of linear combinations with respect to a basis, e.g. (C) = -(D) + 2(A) + 2(B) is the unique way to write (C) as a linear combination of (A), (B), (D), (E) or (A), (B), (D), (F), it is not a conic(al) combination, so no such conic(al) combination exists?

Anyway as hinted above, is this isn't difficult to generalize to the $n \times n$ case, I would love to hear your thoughts about that too. Right now I am trying to understand "the qubit", so $2 \times 2$ case suffices for me.

Related question about minimal spanning sets for conic(al) combinations

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    $\begingroup$ not quite true: $A = (C+D)/2$. $\endgroup$
    – daw
    Nov 28, 2021 at 20:15
  • $\begingroup$ @daw good call, that was actually a typo, $(A)$ shouldn't have been the identity matrix. I wouldn't have noticed that otherwise if you hadn't pointed it out, thank you! $\endgroup$ Nov 28, 2021 at 21:12

1 Answer 1


Consider the matrices $$ \pmatrix{ 1 & t \\ \bar t & t^2 } $$ for $t\in \mathbb C$. All are positive semi-definite. For $|t|>1$ they are no conical combination of your matrices: The entry $t$ fixes one of the matrices C-F, then there is no room to add positive multiples of A,B.

Afaik, this cone cannot be generated by finitely many matrices. So the result you are looking for might not be true.

  • $\begingroup$ To clarify, the reason why no conical combination is possible is because, in order to get a linear combination of those matrices which equals [$1$, $t$, \\ $\bar{t}$, $t^2$], we would need a negative coefficient for [$1$, $0$, \\ $0$, $0$] to "shrink" the upper left entry back down to $1$? Anyway this is a bit of a bummer, but you're right, this does appear to be a valid counterexample. $\endgroup$ Nov 29, 2021 at 16:32

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