19.4F_Final_Builder_Ch_17.3_Least_Squares_$\tkco{eg4}$_$\key{Final}$_Builder_$\…

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The figure below illustrates the relationship between the Im(A)\bct{\text{Im}\, (\A)} and b\bco{\color{cyan}\vb} for the system given by:
{x2y=42x4y=7[  12    24  ][abc] ⁣ ⁣ ⁣ ⁣:=A[xy]=[47][abc]:=b \left\{ \begin{array}{rr} x - 2y \quad =& -4 \\ 2x - 4y \quad =& 7 \end{array} \right. % \qquad \Leftrightarrow \qquad % \underbrace{ \begin{bmatrix} \; 1 & -2 \; \\ \; 2 & -4 \; \end{bmatrix} \vphantom{\colvecth{a}{b}{c}\!\!\!\!} }_{:= \, \bct{\A}} % \colvec{x}{y} % = % \underbrace{ \colvec{-4}{7} \vphantom{\colvecth{a}{b}{c}} }_{:=\, \bco{\color{cyan}\vb}}



Let bIm(A)=projIm(A)(b)\vb_{_{\textrm{Im}(\A)}} = \overrightharpoon{\text{proj}}_{_{\textrm{Im}(\A)}}(\vb), i.e. the (vector) component of b\vb which is in the image of A\A. Identify this vector on the figure; we'll call this component b\vb_{_{||}} for short.

N.B. the vector isn't included in the figure, so by "identify" you'd have to draw it. I'd strongly suggest you make a quick sketch of your own and add the vector to that sketch: this is an essential concept to be comfortable with, so it's worth actually doing.

Let bIm(A)=bb\vb_{\perp_{_{\textrm{Im}(\A)}}} = \vb - \vb_{_{||}}, i.e. b\vb_{\perp} is the (vector) component of b\vb which is not in the image of A\A. Identify this vector on the figure; we'll call this component b\vb_{\perp} for short.
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