Let $X$ be an unknown matrix and $A$ a known matrix, and suppose the problem is well posed in terms of dimensions.
Is there any way to solve the matrix equation $X^TX=A$, analytically or using numerical methods?
Asked
Active
Viewed 316 times
0
-
Is $A$ real or complex? Do $X$ and $A$ have the same dimensions, i.e. $n\times n$, or do you want a low-rank approximation with a smaller $X$? – Chris Culter Sep 24 '19 at 08:22
-
$A$ is real, and they have the same dimensions. – Vinícius Lopes Simões Sep 24 '19 at 08:32
-
1Have you looked at https://math.stackexchange.com/questions/158219/is-a-matrix-multiplied-with-its-transpose-something-special ? It gives some (fairly obvious) properties of $X^TX$. – almagest Sep 24 '19 at 08:41
-
Thanks for the link! I'll read it carefully – Vinícius Lopes Simões Sep 24 '19 at 08:48
1 Answers
0
The matrix $A$ is symmetric. Thus there exists a unitary matrix $U$ such that $UAU^T=D=diag(d_{i})$ is diagonal with real non-negative entries (source: https://en.wikipedia.org/wiki/Symmetric_matrix#Complex_symmetric_matrices).
Thus, $$ A = X^TX = (U^\dagger \sqrt{D})(\sqrt{D}U^*). $$
Inspection tells you that $$ X = \sqrt{D}U^*. $$
Note that this is not unique. For any orthogonal matrix $R$, $\tilde X=RX$ is also a solution.
Daniel
- 195