# 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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if you solve this problem a bit you will get C'BA'=ABC one solution for this equation is C'=A but if you consider the original C'BA'=ABC it has dxd equations with 2xdxb unknowns and for C'=A the number of unknowns is same i.e 2xdxb but number of equations now are dxb. if size of matrix B is greater than D.. then it doesn't seem to be a problem but if its other way round then no of eqns seem to decrease implying there are other equations meaning other solutions. this all is unclear and we not to solve it rigorously for this set of eqns.

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