# Transformation of a symmetrical matrix into another symmetrical matrix

Let $ABC = D$ where $B$ and $D$ are symmetrical matrices. However their [rows x columns] values are not same. For example, $B$ is 2x2 and $D$ is 3x3 a matrix. Clearly, in this case, $A$ has to be a 3x2 matrix and $C$ must be a 2x3 matrix.

Prove or disprove that this holds iff $A^T = C$ i.e. if $A$ and $C$ are transpose to each other then $D$ is symmetrical and if $B$ and $D$ are symmetrical then $A$ and $C$ are transpose to each other. I have tested few cases on computer and it seems to be correct. But I am not sure about 'if and only if' part. Does this hold if $B$ and $D$ are positive definite? Any comment about the nature of $A$ and $C$ is welcome. A chocolate for a correct proof, a cup of coffee otherwise!

EDIT : There was a serious flaw in original problem. Corrected now. Thanks to Hardmath and Willie.

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what exactly is the statement $P$? for example, let $0=A=C^T$ then you can find symmetric matrices $B,D$ such that $ABC=D$ but then $D$ is not of full rank (for the "if" part). –  Prometheus Jan 15 '11 at 12:22
You give as an "example" that $B$ is 2x2 and $D$ is 3x3 as matrices. But if this were the case $D$ could not be full rank, since $ABC$ is rank at most the rank of $B$. –  hardmath Jan 15 '11 at 12:28
@hardmath, [3x2][2x2][2x3] = [3x3] matrix. B is 2x2 and D is 3x3. –  Dilawar Jan 15 '11 at 13:08
@Prometheus. First line is P. I realize that it is ambiguous. Edit. –  Dilawar Jan 15 '11 at 13:09
@Dilawar: if $B$ is $2\times 2$ and rank 2, then the LHS of your expression can have at most rank 2 also. This would contradict the assumption that $D$ is $3\times 3$ and rank 3. –  Willie Wong Jan 15 '11 at 14:26

The claim is untrue, even if we restrict $B,D$ to identity matrices.

A simple example is $(1 0) I (1 1)^T = I$. Clearly $A = (1 0)$ is not the transpose of $C = (1 1)^T$.

In fact $B$ could be any symmetric nxn matrix and $A,C$ any compatible row and column. The result $D$ as a 1x1 matrix will automatically be symmetric.

Added: Examples of larger dimensions are easily constructed as well. For example:

$$A = \begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 0 & 1 \end{pmatrix}$$

$$C = \begin{pmatrix} 1 & 0 \\\\ 1 & 0 \\\\ 0 & 1 \end{pmatrix}$$

will again satisfy $AIC = I$ for identity matrices of compatible dimensions.

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