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?
Necessary and sufficient condition for the matrix $A = I-2xx^t$ to be orthogonal
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0If 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. – 2012-05-10
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0Just multiply $A$ and $A^T$. The formula will give an obvious condition on $x$ so that the product is the identity. – 2012-05-10
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0Request from the down voters: I asked this question long time back so I was unaware of ways of asking questions here. I may delete the question but its not fair as a very nice solution is already given below. – 2017-05-04
2 Answers
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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0As 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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0Then what is $\bf (xx^T)^T=?$ Is it then $\bf (xx^T)^T=xx^T$? – 2014-05-26
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0Thanks a lot..I have got it.I knew the fact $(AB)^T=B^TA^T$ but I just missed it.. – 2014-05-26
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0too calculative @Rahul.. – 2017-06-15
For necessary condition , after calculating $AA^*$ = $I$ we get $XX^*(XX^*- I) = 0$. Now we can say that $XX^*$ , $(XX^* - I)$ are singular matrix. Then we can say $XX^*$ has Eigen value $0$ and $1$ (why?). And we can see that the geometric multiplicity of $0$ is ($n-1$). So algebraic multiplicity of $1$ must be $1$. so trace is $1$. So $X^*X$ is $1$. necessary condition is proved.
now if $X^*X$ is $1$. we know that $XX^*$ has two eigen value one is $0$ with algebraic multiplicity ($n - 1$) and other one is $X^*X$ which is 1 here. so the minimal polynomial is ($s$ - $1$) $s$ = $0$. so we get ($XX^*$ - $1$) $XX^*$ = $0$. we are done..
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1nice but too complicated..@sani – 2017-06-15