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A single Householder similarity reduces a $3 \times 3$ matrix to Hessenberg form

Example

Let

$A = [\begin{array}{ccc} 1 & 2 & 0 \\ 3 & 4 & 5 \\ 4 & 6 & 7 \end{array}], Q = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{3}{5} & \frac{4}{5} \\ 0 & \frac{4}{5} & - \frac{3}{5} \end{array}] .$

Then $Q$ is a Householder reflector and

$Q^{T} A Q = [\begin{array}{ccc} 1 & \frac{6}{5} & \frac{8}{5} \\ 5 & \frac{56}{5} & \frac{3}{5} \\ 0 & - \frac{2}{5} & - \frac{1}{5} \end{array}],$

which is upper Hessenberg.

Facts & Assumptions

Given: The matrix $A$ and reflector $Q$ in the statement.

[L1]

Every real or complex square matrix can be reduced to Hessenberg form by a sequence of Householder similarities (Householder similarities reduce a general matrix to upper Hessenberg form and a real symmetric matrix to tridiagonal form).

Verification

technique · computation

1.1algebra

The lower part of the first column of $A$ is $(3, 4)^{T}$ , whose Euclidean norm is $5$ . The reflector $Q$ fixes the first basis vector and sends $(3, 4)^{T}$ to $(5, 0)^{T}$ .

2.1L1step 1.1algebra∎

Direct multiplication gives the displayed matrix $Q^{T} A Q$ , whose $(3, 1)$ entry is $0$ . Therefore it has no entries below the first subdiagonal, so it is upper Hessenberg. This is the $3 \times 3$ instance of [L1].

Depends on

Householder similarities reduce a general matrix to upper Hessenberg form and a real symmetric matrix to tridiagonal form

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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A single Householder similarity reduces a 3×33×3 matrix to Hessenberg form