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Consider the solution of the equation matrix$A\stackrel{鈫�}{x}=\stackrel{鈫�}{b}$ where$A=\left[\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right]$ and$\stackrel{鈫�}{b}=\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]$ .

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

Substitute the value $\left[\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right]$for A and$\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]$ for$\stackrel{鈫�}{b}$ in the equation${\stackrel{鈫�}{x}}^{*}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈫�}{b}$ .

role="math" localid="1660205424631" $$\begin{gathered} {\stackrel{鈫�}{x}}^{*}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈫�}{b} \\ {\stackrel{鈫�}{x}}^{*}={\left({\left(\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right)}^T\left(\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right)\right)}^{-1}{\left(\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right)}^T\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right] \\ {\stackrel{鈫�}{x}}^{*}={\left(\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right]\right)}^{-1}\left[\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right] \\ {\stackrel{鈫�}{x}}^{*}={\left(\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\right)}^{-1}\left[\begin{array}{c}6\\ 6\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\stackrel{鈫�}{x}}^{*}={\left(\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\right)}^{-1}\left[\begin{array}{c}6\\ 6\end{array}\right] \\ {\stackrel{鈫�}{x}}^{*}={\left(\left[\begin{array}{cc}2/3& 1/3\\ -1/3& 2/3\end{array}\right]\right)}^{-1}\left[\begin{array}{c}6\\ 6\end{array}\right] \\ {\stackrel{鈫�}{x}}^{*}=\left[\begin{array}{c}2\\ 2\end{array}\right] \end{gathered}$$

Therefore, the least square solution of ${\stackrel{鈫�}{x}}^{*}$is$\left[\begin{array}{c}2\\ 2\end{array}\right]$ .

The image of the matrix A is defined as follows.

$$\begin{gathered} lm\left(A\right)=\left\{A\overline{)x}x鈭�{\mathrm{鈩�}}^2\right\} \\ =\left\{\left[\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]\overline{)x}鈭�{\mathrm{鈩�}}^2\right\} \\ lm\left(A\right)=\left\{\left[\begin{array}{c}a+b\\ a\\ b\end{array}\right]\overline{)x}鈭�{\mathrm{鈩�}}^2\right\} \end{gathered}$$

Substitute the value $\left[\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right]$for A , $\left[\begin{array}{c}a+b\\ a\\ b\end{array}\right]$for $lm\left(A\right)$, $\left[\begin{array}{c}2\\ 2\end{array}\right]$for ${\stackrel{鈫�}{x}}^{*}$and $\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]$for $\stackrel{鈫�}{b}$in role="math" localid="1660211257336" ${\left[\stackrel{鈫�}{b}-A{\stackrel{鈫�}{x}}^{*}\right]}^{T}lm\left(A\right)$as follows.

$$\begin{gathered} {\left(\stackrel{鈫�}{b}-A{\stackrel{鈫�}{x}}^{*}\right)}^{T}lm\left(A\right)={\left[\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]-\left[\begin{array}{cc}1& 1\\ 1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}2\\ 2\end{array}\right]\right]}^T\left[\begin{array}{c}a+b\\ a\\ b\end{array}\right] \\ ={\left[\left[\begin{array}{c}3\\ 3\\ 3\end{array}\right]-\left[\begin{array}{c}4\\ 2\\ 2\end{array}\right]\right]}^T\left[\begin{array}{c}a+b\\ a\\ b\end{array}\right] \\ ={\left[\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]\right]}^T\left[\begin{array}{c}a+b\\ a\\ b\end{array}\right] \\ {\left[\stackrel{鈫�}{b}-A{\stackrel{鈫�}{x}}^{*}\right]}^{T}lm\left(A\right)=\left[\begin{array}{ccc}-1& 1& 1\end{array}\right]\left[\begin{array}{c}a+b\\ a\\ b\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\left[\stackrel{鈫�}{b}-A{\stackrel{鈫�}{x}}^{*}\right]}^{T}lm\left(A\right)=\left[\begin{array}{ccc}-1& 1& 1\end{array}\right]\left[\begin{array}{c}a+b\\ a\\ b\end{array}\right] \\ =-a-b+a+b \\ {\left[\stackrel{鈫�}{b}-A{\stackrel{鈫�}{x}}^{*}\right]}^{T}lm\left(A\right)=0 \end{gathered}$$

By the definition of perpendicular of two vectors, the vector$\stackrel{鈫�}{b}-A{\stackrel{鈫�}{x}}^{*}$ is perpendicular to the image of A .

If${\stackrel{鈫�}{x}}^{*}=0$ then it implies$A{\stackrel{鈫�}{x}}^{*}=\stackrel{鈫�}{0}$ .

Hence, the least square solution of ${\stackrel{鈫�}{x}}^{*}$is $\left[\begin{array}{c}2\\ 2\end{array}\right]$and the vector $\stackrel{鈫�}{b}-A{\stackrel{鈫�}{x}}^{*}$is perpendicular to the image of A .