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I need to show that this matrix is orthogonal. I am not familiar enough with vector properties to simplify the right hand side of the equation. $$ H = I - \frac{2}{xx^T}x^Tx $$ where $I$ is the $n\times n$ identity matrix and the $^{T}$s all refer to transposes. Appreciate any help to get started.

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orthogonality means $H^TH = I$, right? We have $$ H^T = I^T - \frac 2{xx^T} (x^Tx)^T = I - \frac 2{xx^T}x^Tx = H $$ so we have to show $H^2 = I$, it holds \begin{align*} H^2 &= I^2 - \frac 4{xx^T}x^Tx + \frac 4{(xx^T)^2} x^Txx^Tx\\\ &= I - \frac 4{xx^T}x^Tx + \frac 4{(xx^T)^2} x^T(xx^T)x\\\ &= I - \frac 4{xx^T}x^Tx + \frac 4{xx^T}x^Tx\\\ &= I. \end{align*}

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We need to prove that $H^TH = I$.

$H^TH = (I-\frac{2}{xx^T}x^Tx)(I-\frac{2}{xx^T}x^Tx) = (I-\frac{4}{xx^T}x^Tx + \frac{4}{(xx^T)^2}(x^Tx)(x^Tx))$

Now I guess x is a row vector. $(x^Tx)(x^Tx) = (x^T(xx^T)x)$ (by associativity of matrix multiplication) $= (xx^T)(x^Tx)$.

So $H^TH = (I-\frac{4}{xx^T}x^Tx + \frac{4}{(xx^T)^2}(xx^T)(x^Tx)) = (I-\frac{4}{xx^T}x^Tx + \frac{4}{(xx^T)}x^Tx) = I$.

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You mean $H=I-\frac{2}{v^T v} v v^T$, I presume, the Householder transformation.

To check for orthogonality, multiply by its transpose $H^T$. A quick check shows that $H=H^T$, so just compute $H H$ and show this is equal to the identity. So, in fact, it is orthonormal.

$$(I-\frac{2}{v^T v} v v^T)(I-\frac{2}{v^T v} v v^T) = I - 2 \frac{2}{v^T v} v v^T + \frac{4}{(v^T v)^2} v v^T v v^T$$

Since $v v^T v v^T = v v^T (v^T v)$, the right hand side reduces to $I$.

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Following copper.hat I replace your $H$ by $H:=I-\frac{2}{v^T v} v v^T$; now the constant vector $v$ is a column vector. I claim that $H$ is the reflection on the orthogonal complement $U$ of the vector $v$.

Proof. When $u\in U$ then $(v\,v^T )u =v(v^T\, u)=0$, whence $Hu=u$, and when $x=\lambda v$ then $Hx=\lambda v-{2\over \|v\|^2}v(v^T\,\lambda v)=-\lambda v =-x$.

This immediately implies that $H$ is orthogonal.

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