# Find $C$, if $A=CBC$, where $A$,$B$,$C$ are symmetric matrices.

If $A=CBC$, where $A$,$B$,$C$ are symmetric matrices and $A$,$B$ are given find $C$.

$A$,$B$,$C$ are assumed to be real valued and $B$ is positive definite matrix. Does the unique solution always exist ? What additional assumptions are needed to obtain unique solution.

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could you please specify in more details what is given and what you want to obtain? –  Fabian Jan 23 '13 at 9:19
You asked for $C$ to be symmetric, but why should $C=ML^{-1}$ be? –  Rahul Jan 23 '13 at 10:25
@Rahul Narain I've edited my question –  Qbik Jan 23 '13 at 16:10

Presumably $A,B,C$ are real symmetric matrices. As $B$ is positive definite, a necessary condition for $A=CBC$ is that $A$ is positive semidefinite. Suppose so. Then $A$ can be orthogonally diagonalized. Therefore, WLOG, we may assume that $A=D\oplus0$ for some real diagonal matrix $D$ with positive diagonal entries. Since $D\oplus0=CBC$, $B$ is positive definite and $C$ is symmetric, $C$ must be of the form $\tilde{C}\oplus0$, where $\tilde{C}$ is a symmetric matrix of the same size as $D$. So, WLOG, we may assume further that $A=D$ is a positive diagonal matrix. Now $A=CBC$ implies that $I = (A^{-1/2}CA^{-1/2})(A^{1/2}BA^{1/2})(A^{-1/2}CA^{-1/2})$. So we may assume further that $A=I$. But then $I=CBC$ means $B=C^{-2}$. That is, $C^{-1}$ is a symmetric square root of $B$. Hence, if $B$ can be orthogonally diagonalized as $B=Q\Sigma Q^T$, then $C=Q\Sigma^{-1/2}\Lambda Q^T$ where $\Lambda$ is any diagonal matrix with each diagonal entry equal to $\pm1$. In other words, $C$ is never unique in this reduced case. Turn back to the original case, we see that a solution $C$ always exists iff $A$ is positive semidefinite and the solution is not unique if $A\not=0$.

Edit: Since positive semidefinite square roots of positive semidefinite matrices are unique, if $C$ is required to be positive semidefinite, the above argument shows that it is uniquely determined by $$C = A^{1/2} (A^{1/2}BA^{1/2})^{-1/2} A^{1/2},$$ where all square roots here are positive semidefinite square roots. (The answer by achille hui gives $C = \sqrt{B}^{-1} \sqrt{ \sqrt{B} A \sqrt{B} } \sqrt{B}^{-1}$. Since $C$ is unique, the two formulae actually produce the same matrix.)

When the assumption that $B$ is positive (or negative) definite is removed, things get nastier. When $B$ is nonsingular and indefinite, $C$ may not be unique even if we impose the condition that $C$ is positive semidefinite. For example, $$\begin{pmatrix}3&-1\\-1&3\end{pmatrix} \begin{pmatrix}1\\&-1\end{pmatrix} \begin{pmatrix}3&-1\\-1&3\end{pmatrix} = (\sqrt{8}I) \begin{pmatrix}1\\&-1\end{pmatrix} (\sqrt{8}I)$$ When $B$ is singular, $C$ simply cannot be unique because its restriction on the kernel of $B$ can be taken to be any self-adjoint operator.

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and could You tell me what would be the most intuitive assumption needed for obtaining unique solution if assumption "B is positive definite matrix" deosn't hold anymore ? –  Qbik Jan 23 '13 at 16:12
@Qbik I think even the most intuitive assumption will involve eigendecompositions of $A,B$ and also the eigenspaces of the product of $C$ with several matrices obtained from the eigendecompositions of $A$ and $B$. So, I don't think there is a simple sufficient condition to guarantee the uniqueness of $C$ in the general case. –  user1551 Jan 24 '13 at 0:54

If $B$ is the identity matrix (which is positive definite as you require), your question becomes whether matrix $A$ has a unique squareroot. Without more assumptions, $A$ may not have a unique squareroot, and in fact an $n \times n$ matrix with $n$ distinct eigenvalues is known to have $2^n$ squareroots.

It is known that a positive definite matrix $A$ has a unique positive definite squareroot, which would be one possible restriction to force uniqueness, but only when the other matrix $B$ happens to be the identity matrix. I don't know what happens for other $B$.

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Given any real symmetric square matrix $M$, it is not hard to show $M$ has a real symmetric square root iff $M$ is positive semi-definite. If $M$ is indeed positive semi-definite, its square root is never unique (unless $M$ is the zero matrix). However, among these square roots, there is one and only one which is positive semi-definite. Let us denote this positive semi-definite square root as $\sqrt{M}$.

Accepting these facts, it is easy to see the equation $A = CBC$ has real symmetric solution for $C$ iff $A$ is positive semi-definite (by looking at the square root of $\sqrt{B} A \sqrt{B}$). When $A$ is indeed positive semi-definite, the "uniqueness" of taking positive semi-definite square root also forces one and only one solution of $C$ to be positive semi-definite. Namely: $$C = \sqrt{B}^{-1} \sqrt{ \sqrt{B} A \sqrt{B} } \sqrt{B}^{-1}$$

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