$A$ is an invertible real square matrix ($A \in \mathbb{M_{n}(\mathbb{R})}$ and $det(A) \neq 0$).
Let's consider another matrix $B \in \mathbb{M_{n}(\mathbb{R})}$ such that: $$B = {}^\intercal A \cdot A$$
Show that $B$ accepts a Cholesky factorization $C \cdot {}^\intercal C$ with $C$ a lower trinagular matrix.
From the first persepctive, I understand that there are exist matrices $D_1$ and $D_2$ such that $D_1 = (D_2)^{-1}$ so $D_1 \cdot D_2 = I_n$ where $I_n$ is the identity matrix.
So, we can write $B = {}^{\intercal} A \cdot A = {}^{\intercal} A \cdot I_n \cdot A $ When we replace $I_n$ with $D_1 \cdot D_2$, we will have:
$$B = {}^\intercal A \cdot D_1 \cdot D_2 \cdot A$$ We choose the matrices $D_1$ and $D_2$ in a way that: $C = {}^\intercal A \cdot D_1$ and ${}^\intercal C = D_2 \cdot A$
I guess that the Gauss elimination matrices should be involved in some way to turn ${}^\intercal A \cdot A$ to the form $C \cdot {}^\intercal C$, but I do not know how they should be related to the matrices $D_1$ and $D_2$ and how to formalize the proof as well.