Two Exercises of Gramian

目录

• Two Exercises of Gramian
  • "Gramian determines shape"
  • "Almost orthonormal"

Two Exercises of Gramian

"Gramian determines shape"

Proposition. 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.

Remark. Similarly, if \(\{v_1,\cdots,v_s\}\) and \(\{w_1,\cdots,w_s\}\) be two subsets of \(\mathbb{C}^n\), then there exists \(A\in U(n)\) 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. We summarize the idea of proving \((\Leftarrow)\) as follows. Given the data \(\langle v_i,v_j \rangle\ (1\le i,j\le s)\), we may focus on a maximal linearly independent subset of \(\{v_1,\cdots,v_s\}\) to study the shape formed by these \(s\) vectors in \(\mathbb{R}^n\). Suppose that \(\{v_{k_1},\cdots,v_{k_r}\}\) is a maximal linearly independent subset of \(\{v_1,\cdots,v_s\}\), then \(\{w_{k_1},\cdots,w_{k_r}\}\) is automatically a maximal linearly independent subset of \(\{w_1,\cdots,w_s\}\). Perform Gram-Schmidt on them simultaneously and extend the resulted orthonormal subsets to two orthonormal bases for \(\mathbb{R}^n\). The associated change of coordinates matrix then fulfills the proof. (The Gram-Schmidt process can be embodied by QR decomposition.)

\((\Rightarrow)\): Obvious.

\[\begin{align*} G(w_1,\cdots,w_s)&=(w_{1}\ \cdots\ w_{s})^{T}(w_{1}\ \cdots\ w_{s})\\ &=(v_{1}\ \cdots\ v_{s})^{T}A^TA(v_{1}\ \cdots\ v_{s})\\ &=(v_{1}\ \cdots\ v_{s})^{T}(v_{1}\ \cdots\ v_{s})=G(v_1,\cdots,v_s)\end{align*} \]

\((\Leftarrow)\): Note that

\[\begin{align*}\text{rank}(v_{1}\ \cdots\ v_{s})&=\text{rank}\, G(v_1,\cdots,v_{s})\\ &=\text{rank}\, G(w_1,\cdots,w_s)=\text{rank}(w_{1}\ \cdots\ w_{s})\end{align*} \]

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,

\[\begin{align*} \text{Null}\,(w_1\ \cdots\ w_r)&=\text{Null}\,(w_1\ \cdots\ w_r)^T(w_1\ \cdots\ w_r)\\ &=\text{Null}\,(v_1\ \cdots\ v_r)^T(v_1\ \cdots\ v_r)=\text{Null}\,(v_1\ \cdots\ v_r)=0 \end{align*} \]

Therefore, there exist \(B,C\in M_{r\times (s-r)}(\mathbb{R})\) such that

\[\begin{align*} (v_1\ \cdots\ v_s)&=(v_1\ \cdots\ v_r)(I_r\ |\ B)\\ (w_1\ \cdots\ w_s)&=(w_1\ \cdots\ w_r)(I_r\ |\ C) \end{align*} \]

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

\[\begin{pmatrix} I_r\\\hline B^T \end{pmatrix}\,G\,(I_r\ |\ B)=\begin{pmatrix} I_r\\\hline C^T \end{pmatrix}\,G\,(I_r\ |\ C)\implies GB=GC\implies B=C \]

Now, it suffices to find some \(A\in O(n)\) such that

\[A(v_1\ \cdots\ v_r)=(w_1\ \cdots\ w_r) \]

By QR decomposition, we have

\[(v_1\ \cdots\ v_r)=Q_1R_1,\quad (w_1\ \cdots\ w_r)=Q_2R_2 \]

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

\[(Q_1R_1)^TQ_1R_1=(Q_2R_2)^TQ_2R_2\implies R_1^TR_1=R_2^TR_2 \]

By the uniqueness of Cholesky decompostion for postive-definite matrices, 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

\[R_1=DR_2,\ R_2^T=R_1^TD\implies D=I_r\implies R_1=R_2 \]

Denote \(R:=R_1=R_2\). Now it suffices to find some \(A\in O(n)\) such that

\[AQ_1=Q_2 \]

Since the columns of \(Q_i\) forms an orthonormal subset of \(\mathbb{R}^n\) and thus extends to an orthonormal basis for \(\mathbb{R}^n\), 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. \(\blacksquare\)

"Almost orthonormal"

The next proposition is one of the problems in my entrance examination of École Polytechnique. It estimates the size of a set of unit vectors that are "almost orthonormal".

Proposition. Show that if \(v_1,\cdots,v_m\) are \(m\) unit vectors in \(\mathbb{C}^n\) such that \(|\langle v_i,v_j \rangle|\le \frac{1}{2\sqrt{n}}\) for any \(i\neq j\), then \(m<2n\).

Proof. Let \(G:=G(v_1,\cdots,v_m)\) be the Gramian, i.e., \(G(i,j)=\langle v_i,v_j \rangle\) for \(1\le i,j\le m\). Define \(H:=G-I_m\). On the one hand, noting that \(H\) is Hermitian, we have

\[\begin{align*} \text{tr}(H^2)&=\text{tr}(H^*H)\\ &=\sum_{i,j=1}^{m}|H(i,j)|^2\\ &=\sum_{1\le i\neq j\le m}|\langle v_i,v_j\rangle|^2\\ &\le (m^2-m)\cdot (\tfrac{1}{2\sqrt{n}})^2=\frac{m^2-m}{4n} \end{align*} \]

On the other hand, since \(G\) is positive semi-definite, \(G\) has \(r:=\text{rank}(G)=\text{rank}(v_1\ \cdots\ v_m)\le n\) positive eigenvalues, and the other \(m-r\) eigenvalues of \(G\) are all zero. (In fact, we only use the latter.) Consequently, the eigenvalues of \(H\) are \(\underbrace{\lambda_1,\cdots,\lambda_r}_{\in (-1,+\infty)},\underbrace{-1,\cdots,-1}_{m-r}\), and the eigenvalues of \(H^2\) are \(\underbrace{\lambda_1^2,\cdots,\lambda_r^2}_{\in [0,+\infty)},\underbrace{1,\cdots,1}_{m-r}\). Therefore,

\[\text{tr}(H^2)=\sum_{i=1}^{r}\lambda_i^2+(m-r)\ge m-n \]

Combining the two estimates, we conclude that

\[m-n\le \frac{m^2-m}{4n} \]

from which \(m<2n\) follows. \(\blacksquare\)