Let \(X\) be an \(n \times 1\) matrix over \(C\) such that \(X^tX = 1\), and \(I_j\) be the \(j\)th column of the identity matrix. Show there is a matrix \(M\) in \(O(n, C)\) such that \(MX = I_j\). If \(X\) has real entries, show there is an \(M\) in \(O(n, R)\) with the property that \(MX = I_j\).
Solution
1. Definition and properties of Householder reflections:
1.1. For any non-zero vector \(w \in F^{n \times 1}\) with non-isotropic norm \(w^t w \neq 0\), define the Householder matrix:\[ H_w = I_n - \frac{2}{w^t w} w w^t. \]
definition of Householder reflection.
1.2. \(H_w\) is symmetric: \(H_w^t = I_n^t - \frac{2}{w^t w} (w w^t)^t = I_n - \frac{2}{w^t w} w w^t = H_w\).
transpose of outer product \((w w^t)^t = w w^t\).
1.3. \(H_w\) is orthogonal (\(H_w \in O(n, F)\)):\[ H_w^t H_w = H_w^2 = \left(I_n - \frac{2}{w^t w} w w^t\right)^2 = I_n - \frac{4}{w^t w} w w^t + \frac{4}{(w^t w)^2} w (w^t w) w^t = I_n. \]
\((w w^t)(w w^t) = w (w^t w) w^t\).
2. Construction of \(M \in O(n, \mathbb{C})\) carrying \(X\) to \(e_j\):
2.1. Let \(X \in \mathbb{C}^{n \times 1}\) with \(X^t X = 1\), and let \(e_j = I_j\) (the \(j\)-th standard basis vector, so \(e_j^t e_j = 1\) and \(e_j^t X = X_j\)).If \(X = e_j\), take \(M = I_n \in O(n, \mathbb{C})\).
\(I_n X = e_j\).
2.2. If \(X \neq e_j\), set \(u = X - e_j\) and \(v = X + e_j\).Compute:\[ u^t u = (X - e_j)^t(X - e_j) = X^t X - 2 e_j^t X + e_j^t e_j = 2(1 - X_j), \]\[ v^t v = (X + e_j)^t(X + e_j) = X^t X + 2 e_j^t X + e_j^t e_j = 2(1 + X_j). \]
expanding bilinear forms.
2.3. Since \(u^t u + v^t v = 4 \neq 0\), at least one of \(u^t u\) or \(v^t v\) is non-zero.
sum of two numbers equals 4.
2.4. Case \(u^t u \neq 0\) (\(X_j \neq 1\)):Define \(M = H_u \in O(n, \mathbb{C})\). Compute:\[ M X = X - \frac{2 u^t X}{u^t u} u = X - \frac{2(X^t X - e_j^t X)}{2(1 - X_j)} (X - e_j) = X - \frac{2(1 - X_j)}{2(1 - X_j)}(X - e_j) = X - (X - e_j) = e_j. \]
1 and 2.
2.5. Case \(v^t v \neq 0\) (\(X_j \neq -1\)):Define \(M = -H_v \in O(n, \mathbb{C})\). Compute:\[ H_v X = X - \frac{2 v^t X}{v^t v} v = X - \frac{2(1 + X_j)}{2(1 + X_j)}(X + e_j) = -e_j \implies M X = -(-e_j) = e_j. \]Since \(-I_n \in O(n, \mathbb{C})\) and \(H_v \in O(n, \mathbb{C})\), \(M \in O(n, \mathbb{C})\).
orthogonal group is closed under multiplication and negation.
3. Real case (\(X \in \mathbb{R}^{n \times 1}\)):
3.1. When \(X \in \mathbb{R}^n\) with \(X^t X = \sum_{k=1}^n X_k^2 = 1\), Cauchy–Schwarz implies \(|X_j| \le 1\).
sum of squares is 1.
3.2. If \(X_j = 1\), then \(\sum_{k \neq j} X_k^2 = 0 \implies X = e_j\), so \(M = I_n \in O(n, \mathbb{R})\) works.
non-negativity of real squares.
3.3. If \(X_j < 1\), then \(u^t u = 2(1 - X_j) > 0\).The Householder matrix \(M = H_u = I_n - \frac{2}{u^t u} u u^t\) has strictly real entries and satisfies \(M \in O(n, \mathbb{R})\) and \(MX = e_j\).
2 step 4 with real vectors.
4. Conclusion:There exists \(M \in O(n, \mathbb{C})\) (and \(M \in O(n, \mathbb{R})\) for real \(X\)) such that \(MX = I_j\). Q.E.D.
2 and 3.