Necessary and sufficient condition for the matrix $A = I - 2 x x^t$ to be orthogonal 
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?

 A: 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 $x$?
A: 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..
