From: Michael Orlitzky <michael@orlitzky.com>
Date: Fri, 30 Nov 2018 16:19:56 +0000 (-0500)
Subject: mjo/cone/symmetric_psd: don't use recusion; work on trivial matrices.
X-Git-Url: http://gitweb.michael.orlitzky.com/?a=commitdiff_plain;h=b4b48ce895557cc5708b090568cfa33e7d09817f;p=sage.d.git

mjo/cone/symmetric_psd: don't use recusion; work on trivial matrices.
---

diff --git a/mjo/cone/symmetric_psd.py b/mjo/cone/symmetric_psd.py
index e75c152..40fd90c 100644
--- a/mjo/cone/symmetric_psd.py
+++ b/mjo/cone/symmetric_psd.py
@@ -46,6 +46,13 @@ def is_symmetric_psd(A):
         sage: is_symmetric_psd(A)
         False
 
+    The trivial matrix in a trivial space is trivially symmetric and
+    positive-semidefinite::
+
+        sage: A = matrix(QQ, 0,0)
+        sage: is_symmetric_psd(A)
+        True
+
     """
 
     if A.base_ring() == SR:
@@ -60,7 +67,11 @@ def is_symmetric_psd(A):
     # semidefinite. For that we can consult its minimum eigenvalue,
     # which should be zero or greater. Since ``A`` is symmetric, its
     # eigenvalues are guaranteed to be real.
-    return min(A.eigenvalues()) >= 0
+    if A.is_zero():
+        # A is trivial... so trivially positive-semudefinite.
+        return True
+    else:
+        return min(A.eigenvalues()) >= 0
 
 
 def unit_eigenvectors(A):
@@ -304,19 +315,23 @@ def random_symmetric_psd(V, accept_zero=True, rank=None):
         # Use the one the user gave us.
         rank_A = rank
 
-    # Begin with the zero matrix, and add projectors to it if we have any.
-    A = V.zero().column()*V.zero().row()
-
-    # Careful, begin at idx=1 so that we only generate a projector
-    # when rank_A is greater than zero.
-    while A.rank() < rank_A:
-        v = V.random_element()
-        A += v.column()*v.row()
-
-    if accept_zero or not A.is_zero():
-        # We either don't care what ``A`` is, or it's non-zero, so
-        # just return it.
-        return A
-    else:
-        # Uh oh, we need to generate a new one.
-        return random_symmetric_psd(V, accept_zero, rank)
+    if n == 0 and not accept_zero:
+        # We're gonna loop forever trying to satisfy this...
+        raise ValueError('You must have accept_zero=True when V is trivial')
+
+    # Loop until we find a suitable "A" that will then be returned.
+    while True:
+        # Begin with the zero matrix, and add projectors to it if we
+        # have any.
+        A = matrix.zero(V.base_ring(), n, n)
+
+        # Careful, begin at idx=1 so that we only generate a projector
+        # when rank_A is greater than zero.
+        while A.rank() < rank_A:
+            v = V.random_element()
+            A += v.column()*v.row()
+
+        if accept_zero or not A.is_zero():
+            # We either don't care what ``A`` is, or it's non-zero, so
+            # just return it.
+            return A