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

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    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?2014-05-25

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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?