<s>[INST] Given $B=AA^T$ and $C=A^TA$, can we calculate $C$ if we know $B$ but not $A$? Consider $B=AA^T$ and $C=A^TA$ with second-order tensor $A$. With knowledge of $B$, but no knowledge of $A$, can we calculate C?
I suspect it should not be possible since different $A$ probably can result in the same $B$, but on the other hand, by the related formulas for $B$ and $C$, maybe all these $A$ also happen to result in the same $C$? How would I go about proving/disproving something like this?
 [/INST] Permute the columns of $A$ doesn't change $B$ but may change $C$, so you can't establish a function $C=f(B)$.
The reason that this is true is that $B_{ij}$ is the dot product of the $i$th and $j$th rows whereas $C_{ij}$ is the dot product of the columns. Permutations on columns have no effect on dot product of rows.
MaxD edited: Starting with
$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
we get
$$B = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
\begin{pmatrix} a & c \\ b & d \end{pmatrix}=
\begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2 \end{pmatrix}$$
and
$$C=\begin{pmatrix} a & c \\ b & d \end{pmatrix}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 
\begin{pmatrix} a^2+c^2 & ab+cd \\ ab+cd & b^2+d^2 \end{pmatrix}.$$
Switching the two columns and starting with
$$A = \begin{pmatrix} b & a \\ d & c \end{pmatrix}$$
results in
$$B = \begin{pmatrix} b & a \\ d & c \end{pmatrix}
\begin{pmatrix} b & d \\ a & c \end{pmatrix}=
\begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2 \end{pmatrix}$$
and
$$C=\begin{pmatrix} b & d \\ a & c \end{pmatrix}
\begin{pmatrix}  b & a \\ d & c \end{pmatrix} = 
\begin{pmatrix} b^2+d^2 & ab+cd \\ ab+cd & a^2+c^2 \end{pmatrix}.$$
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