Question

In, consider the subspace Wspanned by the vectors and $\left[\begin{array}{c}1\\ 1\\ -1\\ 0\end{array}\right]\mathbf{}\mathbf{and}\mathbf{}\left[\begin{array}{c}0\\ 1\\ 1\\ -1\end{array}\right]$. Find the matrixof ${\mathit{P}}_{\mathbf{w}}$the orthogonal projection onto W.

Short answer

The matrix is $P_w=1/3\left[\begin{array}{cccc}1& 1& -1& 0\\ 1& 2& 0& -1\\ -1& 0& 2& -1\\ 0& -1& -1& 1\end{array}\right]$.

Step-by-step solution

Matrix Pwof the orthogonal projection onto W.

Here is need to find the matrix $P_w$of the orthogonal projection onto W.

So, according to the Exercise 67, the required matrix will be$A{\left(A^{T}A\right)}^{-1}{A}^T$, where the matrix .

$A=\left[{\overrightarrow{v}}_1{\overrightarrow{v}}_2\right]$

Then, consider the equations below.

$$\begin{gathered} A^{T}A=\left[\begin{array}{cccc}1& 1& -1& 0\\ 0& 1& 1& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\\ -1& 1\\ 0& -1\end{array}\right] \\ =\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right] \end{gathered}$$

role="math" localid="1660131299371" $$\begin{gathered} A^T{\left(A\right)}^{-1}=1/3\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ A{\left(A^{T}A\right)}^{-1}=1/3\left[\begin{array}{cc}1& 0\\ 1& 1\\ -1& 1\\ 0& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =1/3\left[\begin{array}{cc}1& 0\\ 1& 1\\ -1& 1\\ 0& -1\end{array}\right] \end{gathered}$$

Now $A{\left(A^{T}A\right)}^{-1}{A}^T$is find as,

$$\begin{gathered} A(A^T{A)}^{-1}{A}^T=\left[\begin{array}{cc}1& 0\\ 1& 1\\ -1& 1\\ 0& -1\end{array}\right]\left[\begin{array}{cccc}1& 1& -1& 0\\ 0& 1& 1& -1\end{array}\right] \\ =1/3\left[\begin{array}{cccc}1& 1& -1& 0\\ 1& 2& 0& -1\\ -1& 0& 2& -1\\ 0& -1& -1& 1\end{array}\right] \end{gathered}$$

Hence, the required matrix is $P_w=1/3\left[\begin{array}{cccc}1& 1& -1& 0\\ 1& 2& 0& -1\\ -1& 0& 2& -1\\ 0& -1& -1& 1\end{array}\right]$.