# Thread: Finding a matrix and its transpose from a given matrix

1. Thanks. I will have a try and fix things up before asking for help again. Thanks again JakeD. You have been very helpful

2. Originally Posted by JakeD
Since if $AA^T = B$ implies $B$ is symmetric, there exists an orthogonal matrix $P$ such that $P^TBP = \Lambda,$ and $\Lambda$ is a diagonal matrix with the eigenvalues of $B$ along the diagonal and the columns of $P$ are the eigenvectors of $B$. $P^TBP = \Lambda$ is called a unitary transformation. Since $P$ is orthogonal, $PP^T = I$ and thus $B = PP^T B PP^T = P\Lambda P^T.$

Further, since $x^TBx = x^TAA^Tx = (A^Tx)^T A^Tx > 0$ when $B$ is nonsingular, $B$ is positive definite and thus has positive eigenvalues. Thus $\Lambda = D D^T$ where $D$ has the square roots of the eigenvalues of $B$ on the diagonal.

Then $B = P\Lambda P^T = PD D^TP^T = A A^T$ where $A = PD$. So the problem is reduced to finding the eigenvalues and eigenvectors of a symmetric matrix $B.$ There are efficient numerical methods for this.
Here is the issue when $B = I,$ the identity matrix. It means that $AA^T = I$ and thus $A$ is orthogonal. The problem is that any orthogonal matrix can serve as the eigenvector matrix $P$ for the identity matrix. Thus when the original $A$ is orthogonal, there is no way to identify it by looking at $B = I.$

This brings up a more general point. For any $B,$ a matrix $A$ such that $AA^T = B$ is not necessarily unique. The theory of unitary transformations says nothing about uniqueness.

Page 2 of 2 First 12