Rather than linear algebra and complex numbers u might want to use Groebner bases to solve a quadratic system...Maybe graph theory can help?

Real symmetric matrices are diagonalizable, and it's easy to find the square root of a diagonal matrix. So, first conjugate by a nonsingular matrix P to transform the original matrix into a diagonal one, then replace the diagonal entries with their square roots, and then conjugate that matrix by the inverse of P. The result should be a square root of the original matrix. If the original matrix is not positive-semidefinite (i.e., if it has some negative eigenvalues), it seems like it will not have a real square root.

As already pointed out, the matrix should be diagonalizable. Even if all the roots are positive, resulting in the possibility of a real symmetric solution, you should generally expect 2^N distinct solutions. For each eigenvalue of the original matrix, either the positive or negative square root could be used. (If some of the eigenvalues are zero, there will be fewer than 2^N solutions.

If it's any help, the characteristic polynomial of your big matrix is (lambda-13)^80 x (lambda-256), so your eigenvalues are13,13,13,...,13,256. The 256 comes from 3x(81-1)+16 and the 13 from (16-3).

Ok, I don't think you can get the solution using this method naively. You won't have any way to get rid of the sqrt(13)'s in P\sqrt(D)P^{-1}. An eigenvector of M need not be an eigenvector of sqrt(M), if M has repeated eigenvalues, so not all square roots of M=PDP^{-1} can be written as P\sqrt{D}P^{-1}

U may notice that the square of a matrix aI+bJ (where J is the all-one and I the identity) is a^2I+b(2a+bn)J. To take square root u can identify. This may provide solutions in some cases. Am not sure that all solutions are found that way. For the first problem, n=31,a^2=5, b(2a+bn)=1. Solve for a and b.

I am displaying the 7 by 7 symmetric matrix I want the sqrt of.

3 1 1 1 1 1 1

1 3 1 1 1 1 1

1 1 3 1 1 1 1

1 1 1 3 1 1 1

1 1 1 1 3 1 1

1 1 1 1 1 3 1

1 1 1 1 1 13

Eventually I wish to take sqrt of much larger symmetric matrices for combinatorial research purposesThe answer I am looking for will three 1's in each row and in each column. All other entries will be zero. This answer will be a symmetric binary 7 by 7 matrix.

This is the answer of the last question (7x7), obtained just by playing with combinatorics (it's code theory, at the end, not linear algebra)

1 1 1 0 0 0 0

1 0 0 1 1 0 0

1 0 0 0 0 1 1

0 1 0 1 0 1 0

0 1 0 0 1 0 1

0 0 1 1 0 0 1

0 0 1 0 1 1 0

Yes this is where the question came from. I am attempting to do this with all known symmetric designs.I would like their incidence matrices M so M equals M transpose.

The design (81,16,3) is unresolved. The square of it's incidence matrix will have 16's down the entire diagonal and 3's elsewhere. The square root (if it exists) would yield the incidence matrix of the unresolved design. This is why I am interested in getting the square roots of these larger matrices in an algebraic manner.

Dear Giles

I think that your question has a positive answer ,all what you need is to read about the properties of symmetric matrices (introductory linear algebra) with some knowledge about perumtations and my advise to take care about the upper part of the matrix through computations (your matrix is symmetric) .

regards

Issam: would you do me a favour and take the square root of trhe following matrix

3 1 1 1 1 1 1

1 3 1 1 11 1

1 1 3 1 1 1 1

1 1 1 1 3 1 1

1 1 1 1 1 3 1

1 1 1 1 1 1 3 the answe is a symmetric martix with three 1's in each row and column as shown by Roberto Pignatelli above. The problem we get into is with errors from the floating point arithmetic. Could you help with this by getting a binary solution to the 7 by 7 matrix above. This will put me back on the road to discovery.

Has anyone used Hadamard matrices to find the square roots of symmetric matrices?

Apparently HDHt will work where H is a symmetric Hadamard matrix,Ht is it's transpose,and D is the diagonal matrix.The problem I am having is chaniging the signs on the diagonal entries. With 15 diagonal entries there will be 2^15 possibilities.

A symmetric matrix X will have all real eigenvalues. To find the square root of X, diagonalize it, finding the matrix P such that Inverse(P)*D*P=X, where D is a diagonal matrix with the eigenvalues of X along its diagonal. Now construct E, the diagonal matrix with the square roots of the eigenvalues of X along its diagonal. Note that E^2=D. Then let Y=Inverse(P)*E*P. Then Y^2=X, and so Y=Sqrt(X).

Donald, you're asking for a general solution to the problem of producing a certain special class of SBIBDs. The answer is, in general, unknown, and it's a hard problem. See the CRC Handbook of Combinatorial Designs. Some infinite subclasses are well-known, solved many decades ago.

The way you and others here are describing the problem it sounds as if you believe there is a unique solution in general. There is not. For example, you can expect on the order of n! permutation-equivalent solutions to a problem of this type with nxn matrices. How many inequivalent ones? Again this is hard -- much harder even than the problem of finding one in the first place. The equivalence problem is likely to remain unsolved for good unless some completely new tools are developed.

Rob: Am trying to show that 11 mols of order 12 do not exist so that I can say that the design (157,13,1) does not exist. How would tackle such a problem?

These are simply two well-known formulations for the problem of existence of a projective plane of order 12. If I knew how to successfully tackle this problem I'd likely be famous. The demonstration that there is no projective plane of order 10 took decades, thousands of hours of computation on fast machines with very clever code, and many research papers whittling the problem down to computable size. The order 12 plane is a more distant target. Good luck.

(That said, I can prove that it does not exist, if we require the plane to have certain kinds of internal symmetries. These assumptions allow the use of some high-powered algebraical tools. But it is harder to rule out "sporadic" cases, and that's where the work lies.)

The given matrix

3 1 1 1 1 1 1

1 3 1 1 11 1

1 1 3 1 1 1 1

1 1 1 1 3 1 1

1 1 1 1 1 3 1

1 1 1 1 1 1 3

is a circulant symmetric matrix, therefore its eigen-values and eigen-vectors are known. The matrix can be converted to a diagonal matrix and eigen-values are diagonal elements.Suppose S is the matrix of eigen- vectors where columns are eigen-vectors and D is corresponding diagonal matrix, then inv(S) sqrt(D) S should be the required matrix.

Woulf you kindly send me an example for the 31 by 31 matrix with 6's down the main diagonal and 1's everywhere else. This is M^2 and I want M which is a 31 by 31 binary matrix with 6 ones in each row and in each column. Also the dot product of each pair of rows and cols is 1. Thanks.

@Mittal: what is inv(S)?

Inverse of S.

Could you send me a print out to the solution for a small square symmetric matrix.

Let M^2 be a 21 by 21 matrix with 4's down the diagonal and 1's elsewhere. Now M will be an 21 by 21 binary matrix with 4 ones in each row and column. Also the dot prod of each pair of rows and cols will be 1. Then I will know what to do with larger examples.Thanks so much.

Could you please produce this answer say on a a spreadsheet and send me the answer. This is most important to me. I wish to use it to solve some major problems.

Thanks so much.

Perhaps I may reiterate the difficulty that was pointed out in the first handful of responses to this question, for the recent posters. Indeed, by linear algebra one may diagonalize (over the rationals, since the matrix is symmetric with integer entries) in order to find square roots. There is not one square root, but 2^n, e.g., 2^32 or 2^81, different square roots in the diagonal basis; one must choose one of the two square roots of each of the eigenvalues. (I believe all eigenvalues are nonzero.)

In the original coordinates these matrices have algebraic entries, in the rationals extended by the square roots of the eigenvalues. The original poster is asking for some one of these square roots that is an integer matrix in the original coordinates, or indeed a matrix of only zeros and ones in the original coordinates. (He doesn't say that explicitly, but it seems to be implied.)

Since it appears that these matrices have one eigenvalue of multiplicity n-1, and one can permute the corresponding eigenvectors, it suffices to check 2(n-1) different choices of signs in the diagonal basis, if I'm not mistaken. I'm not a programmer, but It seems to me this might be a manageable computation, once one has diagonalized the original n x n matrix.

since the given matrices are linear combinations of the identify and the matrix commonly denoted J (the all 1 matrix), the eigenvalues are simple to compute (the eigenvalues of the k x k matrix J are one k and the rest 0), but only one of them is a square (36 in the first 31 x 31 example, 256 in the 81 x 81 example), and the rest are all equal (5 in the first example and 13 in the second) - also, since J . J = kJ, a square root of J is J/sqrt(k) - therefore, it is easily shown that a square root of rI + sJ in k x k matrices can be found having the same form aI+bJ, with

r = a^2, s = 2*a+b^2*k, which is easily solved for a and b (need r >= 0 for a to be real, and s-2*a >= 0 for b to be real) - in the two given examples, r=5, s=1 and r=13, s=3, then second condition does not hold and b will be imaginary

there are, of course, other square roots, as other people pointed out