Let u be a unit vector in \({\mathbb{R}^n}\), and let \(B = {\bf{u}}{{\bf{u}}^T}\).

1. Given any x in \({\mathbb{R}^n}\), compute Bx and show that Bx is the orthogonal projection of x onto u, as described in Section 6.2.
2. Show that B is a symmetric matrix and \({B^{\bf{2}}} = B\).
3. Show that u is an eigenvector of B. What is the corresponding eigenvalue?

Consider the following equation shown below:

\(\begin{aligned}{}B{\bf{x}} &= \left( {{\bf{u}}{{\bf{u}}^T}} \right){\bf{x}}\\ &= {\bf{u}}\left( {{{\bf{u}}^T}{\bf{x}}} \right)\\ &= \left( {{{\bf{u}}^T}{\bf{x}}} \right){\bf{u}}\\ &= \left( {{\bf{xu}}} \right){\bf{u}}\\ &= \frac{{{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\end{aligned}\)

Thus, Bx is the orthogonal projection of x onto u.

Apply transpose on both sides of the equation \(B = {\bf{u}}{{\bf{u}}^T}\).

\(\begin{aligned}{}{B^T} &= {\left( {{\bf{u}}{{\bf{u}}^T}} \right)^T}\\ &= {\bf{u}}{{\bf{u}}^T}\\ &= {\bf{B}}\end{aligned}\)

As B is a symmetric matrix, so now check for \({B^2}\).

\(\begin{aligned}{}{B^2} &= B \cdot B\\ &= \left( {{\bf{u}}{{\bf{u}}^T}} \right)\left( {{\bf{u}}{{\bf{u}}^T}} \right)\\ &= {\bf{u}}\left( {{{\bf{u}}^T}{\bf{u}}} \right){{\bf{u}}^T}\\ &= {\bf{u}}{{\bf{u}}^T}\\ &= B\end{aligned}\)

Thus, the equation \({B^2} = B\) is true.

Consider the following equation:

\(\begin{aligned}{}B{\bf{u}} &= \left( {{\bf{u}}{{\bf{u}}^T}} \right){\bf{u}}\\ &= {\bf{u}}\left( {{{\bf{u}}^T}{\bf{u}}} \right)\\ &= {\bf{u}}\left( 1 \right)\end{aligned}\)

Thus, u is the eigenvector of B corresponding eigenvalue 1.