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flip half of matrix over the diagonal to make a symmetric matrix

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Shan Chu
Shan Chu 2016-5-4
评论： Steven Lord 2024-4-9
Dear all, If I have a half of a matrix, e.g
1
2 3
4 5 6
7 8 9 10
...
I want to flip it over the diagonal to make a symmetric matrix:
1 2 4 7
2 3 5 8
4 5 6 9
7 8 9 10
Please help. Thanks
  2 个评论
Jan
Jan 2016-5-4
Tghe solution depends on how the triangular "array" is stored. Are there zeros in the upper right elements?
John D'Errico
John D'Errico 2016-5-4
Is the matrix stored as a matrix, so only the lower triangle, with zeros as the upper triangle. Or is there junk in the upper triangle? Or do you have the elements of the lower triangle, stored in a vector?
All of these things are pertinent to any efficient solution.

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2016-5-4
A=[1 0 0 0
2 3 0 0
4 5 6 0
7 8 9 10]
[n,m]=size(A);
B=A'+A
B(1:n+1:end)=diag(A)
  3 个评论
显示 1更早的评论
Bill Tubbs
Bill Tubbs 2020-5-28
I think you can do it in one line like this:
B = triu(A.',1) + tril(A) % Takes bottom half of A to make B symmetric
Also, this does not do a conjugate transpose.

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更多回答（4 个）

Simon Liljestrand
Simon Liljestrand 2017-9-29
A=[1 0 0 0
2 3 0 0
4 5 6 0
7 8 9 10];
B=A'+triu(A',1)';
Ben McSeveney
Ben McSeveney 2018-2-15
编辑：Stephen23 2018-2-15
If I have a column vector e.g.
1
2
3
How do I quickly create a symmetric matrix i.e.
[1 2 3;
2 1 2;
3 2 1]
?
  3 个评论
显示 1更早的评论
Tom Davis
Tom Davis 2018-2-15
[a,circshift(a,1),circshift(a,2)]
triu(a' - a + ones(size(a,1))) + tril(a - a' + ones(size(a,1))) - eye(size(a,1))

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Walter Bova
Walter Bova 2018-4-16
A = (A+A') - eye(size(A)).*A
Rohit Sachdeva
Rohit Sachdeva 2024-4-9
编辑：Rohit Sachdeva 2024-4-9
As most people have pointed out, I just wanted to add another way of doing this:
B = (A+A') - diag(diag(A));
The (A+A') part is clear to most of us. This is how the 2nd term works:
  • First diag(.) extracts the diagonal elements of A.
  • The next diag(.) creats a matrix with just those diagonal elements.
  • Finally we subtract that matrix of diagonal elements from the (A+A') as required.
This eliminates the need of the eye(.) function. Hope it helps!
  1 个评论
Steven Lord
Steven Lord 2024-4-9
Ran in:
In general, this doesn't work.
A = magic(4)
A = 4x4
16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1
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<mw-icon class=""></mw-icon>
B = (A+A') - diag(diag(A))
B = 4x4
16 7 12 17 7 11 17 22 12 17 6 27 17 22 27 1
<mw-icon class=""></mw-icon>
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It does work if the matrix is real and one of the triangular parts already contains all 0 values.
C = triu(A)
C = 4x4
16 2 3 13 0 11 10 8 0 0 6 12 0 0 0 1
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D = (C+C') - diag(diag(C))
D = 4x4
16 2 3 13 2 11 10 8 3 10 6 12 13 8 12 1
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It doesn't work if the matrix is complex even if the matrix is triangular.
format shortg
C(1, 2) = 2+3i
C =
16 + 0i 2 + 3i 3 + 0i 13 + 0i 0 + 0i 11 + 0i 10 + 0i 8 + 0i 0 + 0i 0 + 0i 6 + 0i 12 + 0i 0 + 0i 0 + 0i 0 + 0i 1 + 0i
D = (C+C') - diag(diag(C))
D =
16 + 0i 2 + 3i 3 + 0i 13 + 0i 2 - 3i 11 + 0i 10 + 0i 8 + 0i 3 + 0i 10 + 0i 6 + 0i 12 + 0i 13 + 0i 8 + 0i 12 + 0i 1 + 0i
D is not symmetric, it is however Hermitian.
issymmetric(D)
ans = logical
0
ishermitian(D)
ans = logical
1
But if you used the non-conjugate transpose then the result is symmetric but not Hermitian:
E = (C+C.')-diag(diag(C))
E =
16 + 0i 2 + 3i 3 + 0i 13 + 0i 2 + 3i 11 + 0i 10 + 0i 8 + 0i 3 + 0i 10 + 0i 6 + 0i 12 + 0i 13 + 0i 8 + 0i 12 + 0i 1 + 0i
issymmetric(E)
ans = logical
1
ishermitian(E)
ans = logical
0

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