Transpose, orthogonal, and inverse

**dwsmith** · April 8th 2010, 06:35 AM

Let u be an unit vector in $R^{n}$ and $H=I-2*u*u^{T}$ . Show that H is orthogonal, symmetric, and its own inverse.

Assume H is orthogonal then $H^{-1}=H^{T}$ and $HH^{T}=I$

$[(I-2*u*u^{T})(I-2*u*u^{T})^{T}]^{T}$

$(I-2*u*u^{T})(I-2*u*u^{T})^{T}$

Now we have $H*H^{T}=I$

I don't think everything is legal here.

**qmech** · April 8th 2010, 09:57 AM

Symmetry:
$H^{T}=(I-2uu^{T})^{T}=I^{T}-2[u^{T}]^{T}u^{T}=I-2uu^{T}=H$

Inverse:
$H^{2}=(I-2uu^{T})*(I-2uu^{T})=I^{2}-4uu^{T}+4uu^{T}uu^{T}$
$=I-4uu^{T}+4uu^{T}=I=HH^{-1}$
remember u is a unit vector.

Orthogonal: Is $H^{T}=H^{-1}$ ?

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