Here we prove that every symmetric matrix is equal to the product A*A^T for some square matrix A.

First we prove the following result.

Lemma 1 If E is obtained from I by applying a row operation p then for every matrix A of the same size as E the product A*E^T coincides with the matrix obtained from A by applying the operation p to the columns of A.

Indeed, by the theorem about transposes we have

Suppose that the operation p deals with rows i and j. Then by the theorem about elementary matrices E*A^T is obtained by applying the operation p to the rows i and j of A^T. These rows are equal to the columns i and j of matrix A. Thus E*A^T is obtained from matrix A by applying operation p to columns i and j and then taking the transpose. Since if we apply the transpose twice nothing changes, (E*A^T)^T is obtained from A by applying the operation p to columns i and j. But we have seen that (E*A^T)^T=A*E^T. The proof is complete.

We shall also need the following observation.

Lemma 2. Let A be a symmetric matrix and p be a row operation. Suppose that we apply p to the rows of A and then apply the same operation to the columns of the resulting matrix. Then the matrix B which we finally get is symmetric and equal EAE^T for some elementary matrix E.

Indeed, let E be the elementary matrix corresponding to the operation p. Then by the theorem about elementary matrices and by our first lemma

Then

We used the theorem about transposes and the fact that A is symmetric (A=A^T). The proof is complete.

Now let us define the following modification of the Gauss-Jordan procedure. Unlike the ordinary Gauss-Jordan procedure, this new procedure applies only to symmetric matrices. Our goal is to get a matrix with 0's and 1's on the diagonal and 0's everywhere else.

Suppose that first j-1 rows of a symmetric matrix A have the form (0,...,0,1,0,...,0) with 1 on the main diagonal or the form (0,0,...,0) and j-th row does not have either of these forms (j may be equal to 1). Then since A is symmetric, the first j-1 columns of A have the same form. This means that A looks like this:

1. If the entry A(j,j) is non-zero, go to step 3 otherwise suppose that the i-th entry of column j, A(i,j), is non-zero, i> ;j. Then add the i-th row of matrix A to the j-th row and the i-th column to the j-th column. By our second lemma the resulting matrix A' is symmetric and A'(j,j) is not zero. So we can go to step 3 now.
2. Multiply the row j and column j by 1/sqrt(A(j,j)). By our second lemma the resulting matrix is symmetric and it is clear that the (j,j)-entry of the first non-zero column of this matrix is 1.
3. If the (j,j)-entry is not 0 for some i> j, then subtract the row j multiplied by A(i,j) from the i-th row and then subtract the column j multiplied by A(i,j) from the i-th column. The resulting matrix will be symmetric and its (i,j)- and (j,i)-entries will be zeros. Thus we can "kill" all non-zero entries in the column j and in the row j. The j-th row (column) of the resulting matrix will have the form (0,...,0,1,0,...,0) with 1 on the main diagonal. If the matrix is still not in the desired form, we can repeat steps 1, 2, 3 until we get a diagonal matrix B with 0's and 1's on the main diagonal.

Thus, let B be the matrix which we obtained from matrix A by our procedure. By lemma 2, we have

for some elementary matrices E₁,...,E_k. Let

Then by the theorem about transposes we have:

By the second theorem about inverses F is invertible. By the first theorem about inverses F^T is invertible and

Thus

Let G=F^-1. Notice that B=B^T, B=B² (since B is diagonal with 0's and 1's on the main diagonal) Therefore

This completes the proof.