Let $A$ and $B$ be two matrices in $M_{n,p}(R)$. Show that $A^{T}A = B^{T}B \iff \exists P \in On(R)$ such that $B = PA$.

Question

let A and B two matrices in $Mn,p(R)$ show that $A^T$$A$=$B^T$$B$ if and only if there exist $P$ belong to $On(R)$ s.t $B$=$PA$.

Answer 1

Hint: Use the fact that $A$ and $B$ have "polar decompositions" $A = U_1P_1$ and $B = U_2P_2$, with $U_1,U_2$ orthogonal and $P_1,P_2$ positive semidefinite.

---

Using the spectral theorem:

We can write $U^T(A^TA)U = D$ for some diagonal $D$ and orthogonal $U$. Let $d_1,\dots,d_n$ denote the diagonal entries of $D$. Define $P = AU$ and $Q = BU$. Note that $P^TP = Q^TQ = D$.

Let $p_i$ denote the $i$th column of $P$ and let $q_i$ denote the $i$th column of $Q$. Using the fact that $P^TP = Q^TQ = D$, we have $$ p_i^Tp_j = q_i^Tq_j = \begin{cases} d_i & i=j \\ 0 & i \neq j. \end{cases} $$ Using this, show that there is an orthogonal matrix $V$ for which $Vp_i = q_i$ for $i = 1,\dots,n$. In other words, $VP = Q$.

Now, show that if $VP = Q$, then $VA = B$.