Problem. Let \(\{v_1,\cdots,v_s\}\) and \(\{w_1,\cdots,w_s\}\) be two subsets of \(\mathbb{R}^n\). Show that there exists \(A\in O(n)\) such that \(Av_i=w_i\ (1\le i\le s)\) iff \(\langle v_{i}, v_{j}\rangle=\langle w_{i}, w_{j}\rangle\ (1\le i,j\le s)\), i.e., the two Gramians are equal.
Proof (Haosen CHEN).
\((\Rightarrow):\)
\((\Leftarrow)\):
Note that
Denote \(r:=\text{rank}\,(v_{1}\ \cdots\ v_{s})\). Without loss of generality, assume that \(v_1,\cdots,v_r\) are linearly independent. We claim that \(w_1,\cdots,w_r\) are linearly independent as well. Indeed,
Therefore, there exist \(B,C\in M_{r\times (s-r)}(\mathbb{R})\) such that
Denote \(G:=G(v_1,\cdots,v_r)=G(w_1,\cdots,w_r)\). Since \(\text{rank}(G)=\text{rank}(v_1\ \cdots\ v_r)=r\), the matrix \(G\) is invertible. Thus we have
Now, it suffices to find some \(A\in O(n)\) such that
By QR decomposition, we have
where \(R_i\in M_{r\times r}(\mathbb{R})\) is an upper triangular matrix with positive digonal entries and \(Q_i\in M_{n\times r}(\mathbb{R})\) satisfies \(Q_i^TQ_i=I_r\) (semi-orthogonal). Thus we have
By the uniqueness of Cholesky decompostion, we have \(R_1=R_2\). Indeed, \(R_1R_2^{-1}=(R_1^{T})^{-1}R_2^T\) is upper triangular and lower triangular, and hence a diagonal matrix, denoted by \(D\). Thus we have
Denote \(R:=R_1=R_2\). Now it suffices to find some \(A\in O(n)\) such that
Since the columns of \(Q_i\) forms an orthonormal subset of \(\mathbb{R}^n\) and thus extends to an orthonormal basis of \(\mathbb{R}^n\), there exists \(\widehat{Q}_i\in O(n)\) such that \(\widehat{Q}_i=(Q_i\ |\ X_i)\) for some \(X_i\in M_{n\times (n-r)}(\mathbb{R})\). Now define \(A:=\widehat{Q}_2\widehat{Q}_1^{-1}\). Then \(A\in O(n)\) and \(AQ_1=Q_2\), as desired. \(\blacksquare\)
原文地址：http://www.cnblogs.com/chaliceseven/p/16853145.html
