OK, I'm working on the following problem and don't understand how my Linear Algebra text has made it to a certain conclusion here is the problem:
Let u be a unit vector in $R^{n}$ and let $H = I - 2\textbf{u}\textbf{u}^{T}$. Show that H is both orthogonal and hence it's own inverse.
So here is the work, I've done and it corresponds with the book, up until a point anyway:
$H^{T} = (I − 2\textbf{u}\textbf{u}^{T})^{T} $
$ = I^{T} − 2(\textbf{u}^{T})^{T} \textbf{u}^{T} $
$ = I − 2\textbf{u}\textbf{u}^{T} = H$
So far so good...
$H^{T}H = H^{2}$
$ = (I − 2\textbf{u}\textbf{u}^{T} )^{2} $
$ = I − 4\textbf{u}\textbf{u}^{T} + 4\textbf{u}\textbf{u}^{T} \textbf{u}\textbf{u}^{T} $
It's at this point the textbook's answer diverges from mine, and states that the previous statement is equal to $ I - 4\textbf{u}\textbf{u}^{T} + 4\textbf{u}\textbf{u}^{T}$. While this is what I want, why is $4\textbf{u}\textbf{u}^{T} \textbf{u}\textbf{u}^{T} = 4\textbf{u}\textbf{u}^{T}$ and not say equal to some value $4(\textbf{u}\textbf{u}^{T})^{2}$?
Your assumption is that $\mathbf{u}$ is a unit vector. This means that $$\mathbf{u}^\mathrm{T}\mathbf{u} = \|\mathbf{u}\|^2 = 1$$ You then have $$4\mathbf{u}\mathbf{u}^\mathrm{T}\mathbf{u}\mathbf{u}^\mathrm{T} = 4\mathbf{u}(\mathbf{u}^\mathrm{T}\mathbf{u})\mathbf{u}^\mathrm{T} = 4\mathbf{u}\|\mathbf{u}\|^2\mathbf{u}^\mathrm{T}=4\mathbf{u}\mathbf{u}^\mathrm{T}$$