# 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

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?

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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

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?