1.8: Projection Matrices
The two-by-two projection matrix projects a vector onto a specified vector in the \(x\)-\(y\) plane. Let \(\mathbf{u}\) be a unit vector in \(\mathbb{R}^2\). The projection of an arbitrary vector \(\mathbf{x} = \langle x_1, x_2\rangle\) onto the vector \(\mathbf{u} = \langle u_1, u_2\rangle\) is determined from
\[\text{Proj}_{\mathbf{u}}(\mathbf{x})=(\mathbf{x}\cdot\mathbf{u})\mathbf{u}=(x_1u_1+x_2u_2)\langle u_1,u_2\rangle .\nonumber \]
In matrix form, this becomes
\[\left(\begin{array}{c}p_1\\p_2\end{array}\right)=\left(\begin{array}{cc}u_1^2&u_1u_2 \\ u_1u_2&u_2^2\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right).\nonumber \]
The projection matrix \(\text{P}_{\mathbf{u}}\), then, can be defined as
\[\begin{aligned}\text{P}_{\mathbf{u}}&=\left(\begin{array}{cc}u_1^2&u_1u_2\\u_1u_2&u_2^2\end{array}\right) \\ &=\left(\begin{array}{c}u_1\\u_2\end{array}\right)\left(\begin{array}{cc}u_1&u_2\end{array}\right) \\ &=\text{uu}^{\text{T}},\end{aligned} \nonumber \]
which is an outer product. Notice that \(\text{P}_{\mathbf{u}}\) is symmetric.
Show that \(\text{P}_{\mathbf{u}}^2=\text{P}_{\mathbf{u}}\).
Solution
It should be obvious that two projections is the same as one. To prove, we have
\[\begin{aligned}\text{P}_{\mathbf{u}}^2&=(\text{uu}^{\text{T}})(\text{uu}^{\text{T}}) \\ &=\text{u}(\text{u}^{\text{T}}\text{u})\text{u}^{\text{T}}&\quad\text{(associative law)} \\ &=\text{uu}^{\text{T}}&\quad (\mathbf{u}\text{ is a unit vector)} \\ &=\text{P}_{\mathbf{u}}.\end{aligned} \nonumber \]