# Recitation 10B

Problem 1: Let y=\begin{bmatrix}4\\8\\1\end{bmatrix}, u_1=\begin{bmatrix}2/3\\1/3\\2/3\end{bmatrix}, u_2=\begin{bmatrix}-2/3\\2/3\\1/3\end{bmatrix}, and W=\mathrm{Span}\{u_1, u_2\}. Let U=\begin{bmatrix}u_1 & u_2\end{bmatrix}. Compute U^TU and UU^T. Compute \mathrm{proj}_Wy and (UU^T)y.

Solution: Compute U^TU=\begin{bmatrix}u_1^T\\u_2^T\end{bmatrix}\begin{bmatrix}u1 & u_2\end{bmatrix}=\begin{bmatrix}u_1^Tu_1 & u_1^Tu_2 \\u_2^Tu_1 & u_2^Tu_2\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}. This implies that u_1, u_2 form an orthornomal basis of W. On the other hand, UU^T = \begin{bmatrix}2/3 & -2/3\\ 1/3 & 2/3\\ 2/3 & 1/3\end{bmatrix}\begin{bmatrix}2/3 & 1/3 & 2/3 \\ -2/3 & 2/3 & 1/3\end{bmatrix}=\frac{1}{9}\begin{bmatrix}8 & -2 & 2 \\ -2 & 5 & 4 \\ 2 & 4 & 5\end{bmatrix}. Because u_1, u_2 form an orthonormal basis, the projection of y onto W is given by (y\cdot u_1)u_1 + (y\cdot u_2)u_2 = \begin{bmatrix}2 \\ 4 \\ 5\end{bmatrix}. Another way to compute the projection is to multiply UU^T with y. In other words, (UU^T)y gives the same vector \begin{bmatrix}2 \\ 4 \\ 5\end{bmatrix}.

Problem 2: Find an orthogonal basis for the column space of the matrix \begin{bmatrix}1 & 2 & 5\\-1 & 1 & -4\\-1 & 4 & -3\\1 & -4 & 7\\1 & 2 & 1\end{bmatrix}.

Solution: Denote its column vectors by x_1, x_2, x_3. Using Gram-Schimidt, we get v_1 = x_1, v_2 = x_2 - \frac{x_2\cdot v_1}{v_1\cdot v_1}v_1 = \begin{bmatrix}3 \\ 0 \\ 3 \\ -3 \\ 3\end{bmatrix}\sim \begin{bmatrix}1 \\ 0 \\ 1 \\ -1 \\ 1\end{bmatrix} and v_3 = x_3 - \frac{x_3\cdot v_1}{v_1\cdot v_1}v_1 - \frac{x_3\cdot v_2}{v_2\cdot v_2}v_2 = \begin{bmatrix}2 \\ 0 \\ 2 \\ 2 \\ -2\end{bmatrix}\sim \begin{bmatrix}1 \\ 0 \\ 1 \\ 1 \\ -1\end{bmatrix}.