# Eigenvalues and eigenvectors of the Householder matrix $H = I - \frac{2}{u^Tu} uu^T$

So during my first revision for the semester exams, I went through exercises in books/internet and I found 2-3 that caught my eye. One of them was the following:

Let $u \in \mathbb R^n$ be a non-zero column vector. Prove that the matrix

$$H = I - \frac{2}{u^Tu}uu^T$$

is symmetric and orthogonal. Then find the eigenvalues and eigenvectors of $H$.

Now first of all, I have already proved that $H$ is symmetric and orthogonal in 2 ways: By definition and by writing the $n$-form of the matrix $H$. After that, I feel that I am lost by trying to calculate the characteristic polynomial of the $n$-form of $H$ and then go the usual way (eigenvalues $\to$ eigenvectors). I am pretty sure I have to work by using the symmetric and orthonormal conditions that I proved first, but I can't get the hang of it. Any tip or help would be appreciated !

I cannot seem to understand why another question with another matrix equation was linked to this one, needless to say, I cannot even understand the answer. I am talking about a differently defined matrix here, with probably different properties and a differently defined question.

• – Jack D'Aurizio May 28 '16 at 16:08
• This is not the same question and your answer personally complicates my mind more that it's already complicated. – Rebellos May 28 '16 at 16:11
• The spectrum is $\{-1,1,1,1,\ldots\}$ and the eigenvectors are given by $u$ and a base of the orthogonal subspace of $u$. Why? That's in my answer. To the other question. – Jack D'Aurizio May 28 '16 at 16:15
• The problem becomes easier if you identify $H$ as a reflection; it reflects a vector with respect to the hyperplane containing $0$ that is normal to $u$. That is, any orthonormal basis extending $u$ will serve as eigenvectors with appropriate eigenvalues. – Sangchul Lee May 28 '16 at 16:22
• Let $v = \dfrac{1}{\| u \|} u$ be the unit vector corresponding to $u$. Then $H = I - 2 v v^T$. This is exactly the equation in the other question, so your question is a duplicate one. – M. Vinay May 28 '16 at 16:24

Given $\mathrm{u} \in \mathbb{R}^n \setminus \{0_n\}$, an eigendecomposition of the projection matrix $\mathrm P := \dfrac{\,\,\mathrm{u} \mathrm{u}^{\top}}{\mathrm{u}^{\top} \mathrm{u}}$ is

$$\mathrm P = \mathrm Q \Lambda \mathrm Q^{\top} = \begin{bmatrix} | & | & & |\\ \dfrac{\mathrm{u}}{\|\mathrm{u}\|} & \mathrm{q}_2 & \cdots & \mathrm{q}_n\\ | & | & & |\end{bmatrix} \begin{bmatrix} 1 & & & \\ & 0 & & \\ & & \ddots & \\ & & & 0\end{bmatrix} \begin{bmatrix} | & | & & |\\ \dfrac{\mathrm{u}}{\|\mathrm{u}\|} & \mathrm{q}_2 & \cdots & \mathrm{q}_n\\ | & | & & |\end{bmatrix}^{\top}$$

where vectors $\mathrm{q}_2, \dots, \mathrm{q}_n$, which can be found using Gram-Schmidt, form an orthonormal basis for the $(n-1)$-dimensional linear subspace orthogonal to $\mathrm{u}$. Hence,

$$\begin{array}{rl} \mathrm H &:= \mathrm I_n - 2 \, \mathrm P = \mathrm I_n - 2 \, \mathrm Q \Lambda \mathrm Q^{\top} = \mathrm Q \mathrm Q^{\top} - 2 \, \mathrm Q \Lambda \mathrm Q^{\top} = \mathrm Q \, (\mathrm I_n - 2 \Lambda)\, \mathrm Q^{\top}\\ &\,\,= \begin{bmatrix} | & | & & |\\ \dfrac{\mathrm{u}}{\|\mathrm{u}\|} & \mathrm{q}_2 & \cdots & \mathrm{q}_n\\ | & | & & |\end{bmatrix} \begin{bmatrix} -1 & & & \\ & 1 & & \\ & & \ddots & \\ & & & 1\end{bmatrix} \begin{bmatrix} | & | & & |\\ \dfrac{\mathrm{u}}{\|\mathrm{u}\|} & \mathrm{q}_2 & \cdots & \mathrm{q}_n\\ | & | & & |\end{bmatrix}^{\top}\end{array}$$

Thus, the eigenvectors of $\mathrm P$ and $\mathrm H$ are the same. The eigenvalues of $\mathrm H$ are

$$\lambda_i (\mathrm H) = 1 - 2 \lambda_i (\mathrm P)$$

The sign of the eigenvalue for the direction of $\mathrm{u}$ has been changed, as $\mathrm P \mathrm{u} = \mathrm{u}$, but $\mathrm H \mathrm{u} = -\mathrm{u}$.

Just express the identity matrix as a product $\mathbf{VV^T}$ where $\mathbf{V}$ is an orthonormal matrix with first column $u/\|u\|$ and the remaining columns some basis of the perpendiculsr space of $u.$ You'll get a matrix decomposition of $H$ showing that the eigenvalues are $(-1,1,..,1),$ eigenvectors are the same as columns of $\mathbf{V}.$