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Let $x$ be a non zeo (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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What have you tried so far? –  Rahul May 10 '12 at 22:46
    
If matrix $A$ is orthogonal then it should satisfy the condition $AA^{t} = I$. Then i came to conclusion that $(XX^{t})^2 = 0$ but it seems to be false. –  srijan May 10 '12 at 22:55
    
Just multiply $A$ and $A^T$. The formula will give an obvious condition on $x$ so that the product is the identity. –  copper.hat May 10 '12 at 23:05

1 Answer 1

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You're right that for $A$ to be orthogonal, you need $AA^T = I$. You may have made a mistake in your derivation. You should get $$AA^T = (I - 2xx^T)(I - 2xx^T) = I - 4xx^T + 4xx^Txx^T = I - 4(1 - x^Tx)(xx^T).$$ In the last step, we use the fact that $x^Tx$ is a scalar and so can be pulled out of the middle of $xx^Txx^T$. So now, for $AA^T$ to equal $I$, either of the two parenthesized terms on the right should be zero. What does this tell you about the vector $x$?

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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? –  learner May 25 at 12:27
    
Then what is $\bf (xx^T)^T=?$ Is it then $\bf (xx^T)^T=xx^T$? –  learner May 26 at 5:32
    
Thanks a lot..I have got it.I knew the fact $(AB)^T=B^TA^T$ but I just missed it.. –  learner May 26 at 10:14

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