Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I asked this question on mathoverflow, but it was deemed too simple, so I'm posting here instead --

Is there a nice way to characterize an orthonormal basis of eigenvectors of the following $d\times d$ matrix?

$$\mathbf{I}-\frac{1}{d} \mathbf{v}\mathbf{v}'$$

Where $\mathbf{v}$ is a $d\times 1$ vector of 1's. This is similar to the Householder matrix, except the $v's$ are not normalized. One eigenvector is $\mathbf{v}$ with corresponding eigenvalue 0, remaining eigenvalues should be 1. I'm looking for an expression in terms of unknown d.

Motivation: this is covariance matrix of uniform multinomial distribution, so expression for orthonormal basis produces a linear transformation that will make variables uncorrelated for large n

Example: below are 5 orthonormal eigenvectors vectors I get from Gram-Schmidt for d=5...what is the expression for general d? An even bigger example -- columns of this form orthonormal basis for d=20



$$-\frac{1}{2 \sqrt{3}},0,\frac{\sqrt{3}}{2},-\frac{1}{2 \sqrt{3}},-\frac{1}{2 \sqrt{3}}$$

$$-\frac{1}{2 \sqrt{5}},\frac{2}{\sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}}$$


Update 09/08 I came across another interesting characterization, when d=2^k, for some k, then Walsh Functions form orthogonal basis for this matrix. In particular, let {$\mathbf{x_i}$} represent the list of vectors of binary expansion of integers 1 to d, ie {(0,0,0),(0,0,1),(0,1,0)...}. Then, rows (and columns) of $M$ define the orthonormal basis of matrix in question, where

$$M_{ij}=(-1)^{x_i \cdot x_j}$$

share|cite|improve this question
hm....formatting seems broken for there standard way to do matrices? (I tried \begin{matrix} and \begin{array}) – Yaroslav Bulatov Sep 7 '10 at 1:35
You need to terminate each row with four backslashes instead of the usual two. – J. M. Sep 7 '10 at 2:22
up vote 3 down vote accepted

If you write $P= I - \frac{1}{d}vv^t$ (I assume your $v'$ means the transpose of $v$), then you certainly have a symmetric matrix

$$ P^t = \left( I- \frac{1}{d}vv^t\right)^t = I -\frac{1}{d}v^{tt}v^t = I - \frac{1}{d}vv^t = P \ . $$

So it diagonalizes and has an orthonormal basis of eigenvectors. That is, there exists an orthogonal matrix $S$, ($S^{-1} = S^t$) and a diagonal matrix $D $ such that $S^tP S = D$.

Moreover, since $P^2 = P$,

\begin{align} P^2 &= \left( I- \frac{1}{d}vv^t\right) \left( I- \frac{1}{d}vv^t\right) \\ &= I - 2\frac{1}{d}vv^t + \frac{1}{d^2}vv^tvv^t \\ &= I - 2\frac{1}{d}vv^t + \frac{1}{d}vv^t \\ &=I- \frac{1}{d}vv^t = P \end{align}

and $P \neq 0,I$, its minimal polynomial is $x^2-x = x(x-1)$. So the only eigenvalues of $P$ are $0,1$.

As for eigenvectors, as you say, $v$ is one of them, of $0$ eigenvalue:

$$ Pv = \left( I- \frac{1}{d}vv^t\right)v = v - \frac{1}{d}vv^tv = v - v = 0 \ . $$

Hence, you can take $\frac{v}{\|v\|} = \frac{1}{\sqrt{d}} (1, \dots ,1)$ as the first vector of your orthonormal basis. The other vectors are an orthonormal basis of the orthogonal complement of $v$, $[v]^\bot$:

\begin{align} \left( I- \frac{1}{d}vv^t\right)w = w &\quad\Longleftrightarrow \quad w - \frac{1}{d}vv^tw = w \\ &\quad\Longleftrightarrow \quad vv^tw = 0 \\ &\quad\Longleftrightarrow \quad v^tw = 0 \end{align}

So, you only need to compute a basis for $[v]^\bot$ and orthonormalize it. In coordinates you must find the solutions of the linear equation

$$ x_1 + \dots + x_d = 0 \ . $$

For instance,

$$ (-1, 1, 0, \dots ,0), (-1, 0, 1, 0, \dots ,0), \dots , (-1, 0, \dots , 0,1) \ . $$

And now you apply the Gram-Schmidt process to these vectors.

EDIT. I think Unkz intuition is right. Before normalizing, the general pattern looks as follows. You have $d-1$ vectors of the form

$$ (-1, 0,\dots,0, i, -1,\dots, -1), \qquad \text{for} \quad i=1,\dots,d-1 \ , $$

where $i$ is in the $d-i+1$ coordinate (so there are $i-1$ coordinates with $-1$ after it) and one more last vector

$$ (d,\dots, d) \ . $$

Let's check that:

(a) Those first $d-1$ vectors belong to $\ker (P-I) = [(1,\dots, 1)]^\bot$ (so they are eigenvectors corresponding to the eigenvalue $1$):

$$ (1,\dots, 1)\cdot (-1, 0,\dots,0, i, -1,\dots, -1) = -1 + i + (i-1)(-1)= 0 \ . $$

(b) Those first $d-1$ are mutually orthogonal vectors. We may assume $i>j$, for instance:

$$ (-1, 0,\dots,0, i, -1,\dots, -1) \cdot (-1, 0,\dots,0, j, -1,\dots, -1) = 1 - i +(i-1)(-1)^2 = 0 \ . $$

Now, you quotient out their norms

$$ \sqrt{1 + i^2 + (i-1)} = \sqrt{i^2 + i} = \sqrt{i (i+1)} \qquad \text{for} \quad i= 1, \dots, d-1 $$


$$ \sqrt{d^2d} $$

and you are done.

share|cite|improve this answer
Well...the last part seems to be the hardest, since d is not known – Yaroslav Bulatov Sep 7 '10 at 1:25

While I haven't taken the time to prove it, if you look at the numbers I think you'll see a very nice pattern if you multiply out all your irrational denominators.

(-1,0,0,0,1) norm=$\sqrt{1 \cdot 2}$

(-1,0,0,2,-1) norm=$\sqrt{2 \cdot 3}$

(-1,0,3,-1,-1) norm=$\sqrt{3 \cdot 4}$

(-1,4,-1,-1,-1) norm=$\sqrt{4 \cdot 5}$

(5,5,5,5,5) norm=$\sqrt{5^2 \cdot 5}$

This appears to work for a few other random choices of d, where you have -1 in the first column, (1..d) down the backwards diagonal, and -1 under the backwards diagonal, along with 1,1,...,1 on the bottom row.

share|cite|improve this answer

One way is to find the householder matrix Q that maps v to a multiple of e_1 (first coordinate basis vector). Then (since Q is symmtric and orthogonal) w_2=Q*e_2 ... will be a basis of the orthogonal complement of v. Explicitly, I get w_k = e_k - u where u = (v+sqrt(d)*e_1)/(d+sqrt(d))

share|cite|improve this answer

a slight retouching of unkz's idea is the following orthogonal basis for $v^\perp$:

$$u_1 = (1,-1,0,\dots,0),\ u_2 = (1,1,-2,0,\dots,0),\ \dots, \ u_{d-1} = (1,\dots,1,-d+1)$$

in this form it is pretty easy to see that the $\{u_i\}$ are in $v^\perp$ and are mutually orthogonal.

i don't quite understand the remark that normalizing the $\{u_i\}$ to get an ONB is the hardest part - since $d$ is unknown. what sort of expression "in terms of [the] unknown $d$" would be suitable?

ps: i would post this as a comment - if this @!*&*%#! site would let me.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.