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?
 A: No. Consider matrices of the form $A = \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix}$ for example.
A: Thinking about the SVD should immediately make you suspicious of this claim: writing $A=U\Sigma V^T$, notice that when you compute $B = AA^T = U\Sigma^2 U^T$, you "throw away" all of the information contained in $V$. Since $V$ appears in $C = V \Sigma^2 V^T$, it's intuitive that changing $V$ keeps $B$ the same but generally will change $C$, so that it's impossible to recover $C$ from only $B$.
This is not a proof, since as you say, it's possible that even though $V$ appears in the formula for $C$, it could turn out that $C$ is invariant with respect to changing $V$. Indeed, in the case where $\Sigma = I$ is the identity matrix, then $C = VV^T = I$ does not depend on $V$.
Nevertheless, the intuition should lead you on the path to a proof. Suppose $A$ contains a pair of unequal singular values. Can you show that $C$ can change depending on the value of $V$, so that it's impossible to reconstruct $C$ from only $B$?
A: 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
$$C=\begin{pmatrix} b & d \\ a & c \end{pmatrix}
\begin{pmatrix}  b & a \\ d & c \end{pmatrix} = 
\begin{pmatrix} b^2+d^2 & ab+cd \\ ab+cd & a^2+c^2 \end{pmatrix}.$$
A: If we replace $A$ with $A R$ where $R$ is a unitary (rotation) matrix, we would get: $$B = A R (AR)^T = AR  R^T A^T = AA^T$$
as a unitary (real) matrix is defined by $RR^T = I$. But under these new matrices $C$ becomes $R^T A^T A R$ . So $B$ does not uniquely determine $C$.
It does however determine $C$ up to unitary transformation (ie. you will know the eigenvalues of $C$,as they are always equal to the eigenvalues of $B$ but you will not know $C$. This is provable using singular value decomposition.
