Question

In R^4, let U=span((1,1,0,0),(1,1,1,2)). Find u$\in$U such that ||u-(1,2,3,4)|| is as small as possible.

Answer 1

$\text{We will use the normal equations for the formula of an orthogonal projection. Let}\\A=\begin{pmatrix} 1 & 1 \\1&1\\0&1\\0&2 \end{pmatrix} and b = \begin{pmatrix} 1 \\2\\3\\4 \end{pmatrix} \\\text{Then u is the orthogonal projection of b onto the subspace spanned by the column }\\\text{of A and it is given by the formula:}\\u= A(A^TA)^{-1}A^Tb\\\text{We get that }\\A^Tb =\begin{pmatrix} 1 & 1&0&0 \\1&1&1&2 \end{pmatrix}\begin{pmatrix} 1\\2\\3\\4 \end{pmatrix}=\begin{pmatrix} 3\\14 \end{pmatrix}\\ A^TA=\begin{pmatrix} 1 & 1&0&0 \\1&1&1&2 \end{pmatrix}\begin{pmatrix} 1 & 1 \\1&1\\0&1\\0&2 \end{pmatrix}=\begin{pmatrix} 2&2\\2&7 \end{pmatrix} \\(A^TA)^{-1}=\frac{1}{10} \begin{pmatrix} 7 & -2 \\ -2 & 2 \end{pmatrix}\\(A^TA)^{-1}A^Tb=\frac{1}{10} \begin{pmatrix} 7 & -2 \\ -2 & 2 \end{pmatrix}\begin{pmatrix}3\\14\end{pmatrix}=\frac{1}{10} \begin{pmatrix} -7 \\ 22 \end{pmatrix} \\u=A(A^TA)^{-1}A^Tb=\frac{1}{10} \begin{pmatrix} 1&1 \\ 1&1\\0&1\\0&2 \end{pmatrix}\begin{pmatrix}-7\\22\end{pmatrix}=\frac{1}{10} \begin{pmatrix} 15 \\ 15\\22\\44 \end{pmatrix} \\\text{The answer is u=} \frac{1}{10}\begin{pmatrix}15\\15\\22\\44\end{pmatrix}$