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

linear-algebra matrices eigenvalues-eigenvectors orthogonal-matrices

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?


Solution 1:

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 $x$?

Related

Orbit space of torus homeomorphic to mobius strip
If $ \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $, then is it true that $ \lim_{n \to \infty} x_{n} = x $?
Power series representation/calculation
Maximal inequality for a sequence of partial sums of independent random variables [closed]
variance inequality
Partition an integer $n$ into exactly $k$ distinct parts
If $ab \equiv r \pmod{p}$, and $x^2 \equiv a \pmod{p}$ then $y^2 \equiv b \pmod{p}$ for which condition of $r$?
Matrix commuting with commutator
Prove that an orbit is heteroclinic
$\int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)$ [duplicate]
Extending a morphism of schemes
Show that that $|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}$