Constructing a symmetric matrix

4 次查看（过去 30 天）
A
A 2015-3-13
评论： Andrei Bobrov 2015-3-13
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!

请先登录，再回答此问题。

回答（1 个）

Roger Stafford
Roger Stafford 2015-3-13
编辑：Roger Stafford 2015-3-13
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;" ]

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by