TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-30

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Householder reflectors and Givens transformations are unitary and can annihilate prescribed entries

Statement

Let $F = R$ or $C$ .

Every Householder reflector $H_{v} = I - 2 v v^{*}$ is orthogonal or unitary.
For every nonzero $z \in F^{m}$ , there is a Householder reflector $H$ such that $H z = α e_{1}$ with $∣ α ∣ = ∥ z ∥_{2}$ .
For every pair $(a, b) \in F^{2}$ not both zero, there is a real or complex Givens transformation $G$ such that $G (\binom{a}{b}) = (\binom{α}{0})$ with $∣ α ∣ = (∣ a ∣^{2} + ∣ b ∣^{2})^{1 / 2}$ .

Facts & Assumptions

Given: A field $F = R$ or $C$ , a unit vector $v$ , a nonzero vector $z \in F^{m}$ , and a pair $(a, b) \in F^{2}$ with not both entries zero.

[L1]

Householder reflectors and Givens transformations are defined in Householder reflectors in real or complex inner-product spaces and Real Givens rotations and complex Givens transformations.

[L2]

An orthogonal or unitary operator is an invertible linear isometry (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

Proof

technique · direct matrix multiplication and explicit parameter choice

1.1L1L2algebra

For $H_{v} = I - 2 v v^{*}$ with $v^{*} v = 1$ , one has $H_{v}^{*} = I - 2 v v^{*} = H_{v}$ and $H_{v}^{2} = I - 4 v v^{*} + 4 v (v^{*} v) v^{*} = I$ . Hence $H_{v}^{*} H_{v} = I$ , so $H_{v}$ is orthogonal or unitary.

1.2L1algebra

In the real case, for $(a, b) \neq (0, 0)$ set $r = (a^{2} + b^{2})^{1 / 2}$ , $c = a / r$ , and $s = b / r$ . Then $c^{2} + s^{2} = 1$ and $(\begin{array}{cc} c & s \\ - s & c \end{array}) (\binom{a}{b}) = (\binom{r}{0}) .$ In the complex case, if $a = 0$ take $c = 0$ and $s = b / ∣ b ∣$ ; otherwise write $a = ϕ ∣ a ∣$ with $∣ ϕ ∣ = 1$ , set $r = (∣ a ∣^{2} + ∣ b ∣^{2})^{1 / 2}$ , $c = ∣ a ∣ / r$ , and $s = ϕ \overline{b} / r$ . Then $c^{2} + ∣ s ∣^{2} = 1$ and $(\begin{array}{cc} c & s \\ - \overline{s} & c \end{array}) (\binom{a}{b}) = (\binom{ϕ r}{0}) .$ Thus a Givens transformation annihilates the second entry while preserving the Euclidean norm.

2.1L1step 1.1algebra

If $z$ is already a scalar multiple of $e_{1}$ , choose any Householder reflector fixing $e_{1}$ : for $m = 1$ , $H = [- 1]$ works; for $m > 1$ , take $v = e_{2}$ , so $H_{v} e_{1} = e_{1}$ . Otherwise set $α : = e^{i θ} ∥ z ∥_{2}$ , where $e^{i θ} = z_{1} / ∣ z_{1} ∣$ when $z_{1} \neq 0$ and $e^{i θ} = 1$ when $z_{1} = 0$ , and put $w : = z - α e_{1}$ , $v : = w / ∥ w ∥_{2}$ . Then $v^{*} z = (∥ z ∥_{2}^{2} - \overline{α} z_{1}) / ∥ w ∥_{2} = ∥ w ∥_{2} / 2$ , so $H_{v} z = z - 2 v (v^{*} z) = z - w = α e_{1}$ .

3.1step 1.1step 2.1step 1.2∎

Step 1.1 proves claim 1, step 2.1 proves claim 2, and step 1.2 proves claim 3.

Depends on

Used by

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Sources

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Theorem 3.4.1 and Givens rotations section (standard reference, not scraped)
David Bindel, CS 4220: Numerical Analysis, Least squares and QR (standard reference, not scraped)