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

Let x be a non-zero (column) vector in $R^n$. What is the necessary and sufficient condition for the matrix $A = I − 2xx^T$ to be orthogonal?

share|cite|improve this question
what does "orthogonal" matrix mean? – James S. Cook Sep 1 '12 at 18:01
We have $AA^T = (I − 2xx^T)(I − 2 x x^T)^T = (I − 2xx^T)(I − 2 x x^T) = I -4xx^T + 4xx^T xx^T$. Let $X = xx^T$. Then $AA^T = I$ iff $X = X^2$, or $X$ has the minpoly $\lambda^2 - \lambda$. (edit: better use Tapu's answer). That is, the condition is: $X = \| x \|^2 X$ where $X \neq 0$. – user2468 Sep 1 '12 at 18:08
As I see that $AA^T=(I-2 \bf xx^T)(I-2 \bf xx^T)^T=(I-2 \bf xx^T)(I-2 \bf x^Tx)=I-2\bf x^Tx-2xx^T+4xx^Tx^Tx$. Now,$\bf x^Tx=xx^T$ has been used to reach the result $AA^T=I-\bf 4 xx^T+ 4xx^Txx^T$ Sorry for my dumbness. I fail to understand why $\bf xx^T=x^Tx$ as $\bf x$ being a non-zero vector in $\Bbb R^n$,will be a $n \times 1$ matrix where as $\bf x^T$ will be $1 \times n$ matrix. So, $\bf xx^T$ is a $n \times n$ matrix whereas $\bf x^Tx$ is a $1 \times 1$ matrix. Can someone please explain? – learner May 25 '14 at 12:27
up vote 1 down vote accepted

Hint: $A$ is orthogonal if and only if $A.A^T=I$. Note that in your case $A^T=A$ and $x^Tx=\|x\|^2$. So, after some simplification, we have $AA^T=I+4(\|x\|^2-1)xx^T$. when the quantity on the right hand side be I?

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.