Are the product and sum of two Householder matrices also Householder matrices?

Question

Suppose we have two Householder matrices

$$H_i=I_n-2v_iv_i^T$$

where $\|v_i\|_2 = 1$ for $i \in \{1,2\}$.

• Is it true that $H_1 H_2$ is also a Householder matrix?

• Is it true that $H_1 + H_2$ is also a Householder matrix?

Answer 1

• Negative answer for the product:

The reason is that a Householder matrix has determinant $-1$.

Proof: a Householder transformation is a reflection with respect to an hyperplane with equation $V^TX=0$, its eigenvalues are $-1$ (corresponding eigenvector $V$) and $1$ (corresponding eigenvectors: any base of the hyperplane) with respective multiplicities $1$ and $n-1$.

Thus the product of two Householder matrices has determinant $+1$.

• Negative answer for the sum:

Case $n \geq 3$: this is due to the following fact. If $H_1$ and $H_2$ denote the two hyperplanes, $H_1 \cap H_2$ has dimension $n-2$. Let us take any $Y\neq 0 \in H_1 \cap H_2$ ; were $H_1+H_2$ an isometry, we should have:

$$\|Y\| = \|(H_1+H_2)Y\|=\|H_1Y+H_2Y\| = \|Y+Y\|= 2\|Y\| $$

Contradiction.

The case $n=2$ can be shown impossible as well, because of the structure of reflection matrices, which is:

$$\tag{1}\begin{pmatrix}a & \ \ b\\b & -a\end{pmatrix} \ \text{with the constraint} \ a^2+b^2=1.$$

The sum of two such matrices

$$\begin{pmatrix}a_1 & \ \ b_1\\b_1 & -a_1\end{pmatrix}+\begin{pmatrix}a_2 & \ \ b_2\\b_2 & -a_2\end{pmatrix}$$

gives an impossibility: we cannot have in general: $a_1^2+b_1^2=1$ and $a_2^2+b_2^2=1$ together with $(a_1+a_2)^2+(b_1+b_2)^2=1.$ unless the dot product $a_1a_2+b_1b_2$ is zero.

Answer 2

The product is

$$\rm H_1 H_2 = \left( I_n - 2 v_1 v_1^{\top} \right) \left( I_n - 2 v_2 v_2^{\top} \right) = I_n - 2 v_1 v_1^{\top} - 2 v_2 v_2^{\top} + 4 v_1 v_1^{\top} v_2 v_2^{\top}$$

If $\rm v_1, v_2$ are orthonormal, then $\rm v_1^{\top} v_2 = 0$ and, thus,

$$\rm H_1 H_2 = I_n - 2 v_1 v_1^{\top} - 2 v_2 v_2^{\top} = I_n - 2 \, \underbrace{\begin{bmatrix} | & |\\ \mathrm v_1 & \mathrm v_2\\ | & |\end{bmatrix}}_{=: \mathrm V} \begin{bmatrix} — \mathrm v_1^{\top} —\\ — \mathrm v_2^{\top} —\end{bmatrix} = I_n - 2 V V^{\top}$$

is quasi-Householder as it is symmetric, orthogonal and involutory. Note that $\rm V V^{\top}$ is a projection matrix (because $\rm v_1, v_2$ are orthonormal, i.e., $\rm V^{\top} V = I_2$).

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Suppose we are given a symmetric matrix $\mathrm A \in \mathbb R^{n \times n}$. Let its spectral decomposition be

$$\rm A = Q \Lambda Q^{\top} = \begin{bmatrix} \mathrm Q_1 & \mathrm Q_2\end{bmatrix} \begin{bmatrix} \Lambda_1 & \\ & \Lambda_2\end{bmatrix} \begin{bmatrix} \mathrm Q_1^{\top}\\ \mathrm Q_2^{\top}\end{bmatrix}$$

Suppose further that $\rm Q_1 = V$. Thus, the spectral decomposition of $\rm H_1 H_2$ is

$$\rm H_1 H_2 = \begin{bmatrix} \mathrm Q_1 & \mathrm Q_2\end{bmatrix} \begin{bmatrix} -\mathrm I_2 & \\ & \mathrm I_{n-2}\end{bmatrix} \begin{bmatrix} \mathrm Q_1^{\top}\\ \mathrm Q_2^{\top}\end{bmatrix}$$

Hence,

$$\rm (H_1 H_2) A = \begin{bmatrix} \mathrm Q_1 & \mathrm Q_2\end{bmatrix} \begin{bmatrix} -\Lambda_1 & \\ & \Lambda_2\end{bmatrix} \begin{bmatrix} \mathrm Q_1^{\top}\\ \mathrm Q_2^{\top}\end{bmatrix}$$

Left-multiplying matrix $\rm A$ by the quasi-Householder matrix $\rm H_1 H_2$ reverses the sign of the first two eigenvalues of $\rm A$ (the ones corresponding to eigenvectors $\rm v_1, v_2$), while preserving all eigenvectors.