Suppose I've got an $n \times n$ invertible matrix $A$, and any vector $\mathbf{x}$. Suppose the following identity holds true: $$ A^{T}A - \mathbf{x} \mathbf{x}^{T} = \mathbb{I} $$
Is it then true that $AA^{T} - \mathbf{x}^{T} \mathbf{x} \mathbb{I} = \mathbb{I}$?
I'm wondering this because of the Sylvester determinant identity.
When $n>1$ and $x\ne0$, the answer is always false, regardless of the underlying field. We don't even need to assume that $A$ is invertible; we only need $A$ to be a square matrix.
Suppose the contrary that $AA^T-x^TxI=I$. Since $A^TA$ and $AA^T$ have identical characteristic polynomials, $A^TA=I+xx^T$ and $AA^T=(1+x^Tx)I$ have identical spectra. However, all eigenvalues of $(1+x^Tx)I$ are equal to $1+x^Tx$, while $I+xx^T$, being a rank-1 update to the identity matrix, must have at least $n-1$ eigenvalues equal to $1$. Therefore $1+x^Tx=1$ and $AA^T=I$. But then $A^TA=AA^T$ and in turn $I+xx^T=I$, which is impossible because $x\ne0$.