Question

Find the least-squares solution $\stackrel{\mathbf{\to }\mathbf{*}}{\mathbf{x}}$of the system $\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{b}}$, where$\mathit{A}\mathbf{=}\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]$ and$\overrightarrow{\mathbf{b}}\mathbf{=}\left[\begin{array}{c}1\\ -2\\ 3\end{array}\right]$. Explain

Short answer

The least square solution of$\stackrel{\to *}{x}$ is$\left[\begin{array}{c}0\\ 0\end{array}\right]$ and the vector$\overrightarrow{b}$ is perpendicular to the column of A.

Step-by-step solution

Determine the least squares solution.

Consider the solution of the equation matrix$A\overrightarrow{x}=\overrightarrow{b}$where$A=\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]$and $\overrightarrow{b}=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$.

If is the solution of the equation$A\overrightarrow{x}=\overrightarrow{b}$then the least-square solution$\stackrel{\to *}{x}$is defined as$\overrightarrow{x}={\left(A^{T}A\right)}^{-1}{A}^T\overrightarrow{b}$.

Substitute the value $\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]$for A and $\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]$for in the equation $\stackrel{\to *}{x}={\left(A^{T}A\right)}^{-1}{A}^T\overrightarrow{b}$.

$\stackrel{\to *}{x}={\left(A^{T}A\right)}^{-1}{A}^T\overrightarrow{b}$

role="math" localid="1660126240637" $$\begin{gathered} \stackrel{\to *}{x}={\left({\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]}^T\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]\right)}^{-1}{\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]}^T\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right] \\ \stackrel{\to *}{x}={\left(\left[\begin{array}{ccc}1& 2& 1\\ 1& 8& 5\end{array}\right]\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]\right)}^{-1}\left[\begin{array}{ccc}3& 5& 4\\ 2& 3& 5\end{array}\right]\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right] \\ \stackrel{\to *}{x}={\left(\left[\begin{array}{cc}6& 22\\ 22& 90\end{array}\right]\right)}^{-1}\left[\begin{array}{c}68\\ 47\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

role="math" localid="1660126334311" $$\begin{gathered} \stackrel{\to *}{x}={\left(\left[\begin{array}{cc}6& 22\\ 22& 90\end{array}\right]\right)}^{-1}\left[\begin{array}{c}68\\ 47\end{array}\right] \\ \stackrel{\to *}{x}=\left[\begin{array}{cc}90/28& -22/28\\ -22/28& 6/28\end{array}\right]\left[\begin{array}{c}68\\ 47\end{array}\right] \\ \stackrel{\to *}{x}=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

Therefore, the least square solution of$\stackrel{\to *}{x}$ is $\left[\begin{array}{c}0\\ 0\end{array}\right]$.

Explain the meaning of x→*=0 .

If$\stackrel{\to *}{x}=0$implies $A\stackrel{\to *}{x}=\overrightarrow{0}$.

As $A\stackrel{\to *}{x}$is orthogonal projection of $\overrightarrow{b}$onto $Im\left(A\right)$and$A\stackrel{\to *}{x}=\overrightarrow{0}$ means the vector$\overrightarrow{b}$ is perpendicular to the column of A.

Hence, the least square solution of $\stackrel{\to *}{x}$is $\left[\begin{array}{c}0\\ 0\end{array}\right]$and the vector $\overrightarrow{b}$is perpendicular to the column of A .