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Constructing a symmetric matrix

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A
A on 13 Mar 2015
Commented: Andrei Bobrov on 13 Mar 2015
Hello!
I am trying to construct a symmetric matrix, B, from a random matrix, A
My code is as follows:
scale=6;
A = randn(scale,scale);
%for real matrices, unitary is the same as orthogonal
[Q,R]=qr(A);
%Q is a unitary matrix, R is an upper-triangular matrix. A=Q*R
%construct a set of equally spaced values, d, from which a diagonal matrix is made
d=[1:1:scale];
D=diag(d);
%A symmetric matrix can be constructed from Q*D*transpose(Q)
B=Q*D*transpose(Q) %symmetric matrix
tf=issymmetric(B) %but it isn't symmetrical!
I think the problem arises from there being negative values from Q*transpose(Q), it isn't a perfectly unitary matrix. This is the output from the above code for Q*transpose(Q):
1.0000 -0.0000 0.0000 0.0000 0.0000 -0.0000
-0.0000 1.0000 0.0000 -0.0000 -0.0000 0.0000
0.0000 0.0000 1.0000 -0.0000 0.0000 -0.0000
0.0000 -0.0000 -0.0000 1.0000 -0.0000 0.0000
0.0000 -0.0000 0.0000 -0.0000 1.0000 0.0000
-0.0000 0.0000 -0.0000 0.0000 0.0000 1.0000
So, something needs to be fixed in my step to create the symmetric matrix, B, but I'm not exactly sure what. Any advice would be greatly appreciated!

Answers (1)

Roger Stafford
Roger Stafford on 13 Mar 2015
Edited: Roger Stafford on 13 Mar 2015
Very likely your B matrix only lacks symmetry because of round-off errors in the computation process. Do this instead of using 'issymmetric', which demands exact symmetry:
tf = max(max(abs(B-B.')))<tol;
for some very small tolerance value, 'tol' - that is, small relative to the values in the original Q matrix.
[Note added: If you still need exact symmetry after passing the above test, you can write "B = (B+B.')/2;" ]

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