1.8: Matrices de Proyección
- Page ID
- 119148
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)La matriz de proyección de dos por dos proyecta un vector sobre un vector especificado en el\(y\) plano\(x\) -. Dejar\(\mathbf{u}\) ser un vector de unidad en\(\mathbb{R}^2\). La proyección de un vector arbitrario\(\mathbf{x} = \langle x_1, x_2\rangle\) sobre el vector\(\mathbf{u} = \langle u_1, u_2\rangle\) se determina a partir de
\[\text{Proj}_{\mathbf{u}}(\mathbf{x})=(\mathbf{x}\cdot\mathbf{u})\mathbf{u}=(x_1u_1+x_2u_2)\langle u_1,u_2\rangle .\nonumber \]
En forma de matriz, esto se convierte
\[\left(\begin{array}{c}p_1\\p_2\end{array}\right)=\left(\begin{array}{cc}u_1^2&u_1u_2 \\ u_1u_2&u_2^2\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right).\nonumber \]
La matriz de proyección\(\text{P}_{\mathbf{u}}\), entonces, se puede definir como
\[\begin{aligned}\text{P}_{\mathbf{u}}&=\left(\begin{array}{cc}u_1^2&u_1u_2\\u_1u_2&u_2^2\end{array}\right) \\ &=\left(\begin{array}{c}u_1\\u_2\end{array}\right)\left(\begin{array}{cc}u_1&u_2\end{array}\right) \\ &=\text{uu}^{\text{T}},\end{aligned} \nonumber \]
que es un producto externo. Observe que\(\text{P}_{\mathbf{u}}\) es simétrico.
\(\text{P}_{\mathbf{u}}^2=\text{P}_{\mathbf{u}}\)Demuéstralo.
Solución
Debe ser obvio que dos proyecciones es igual que una. Para probarlo, tenemos
\[\begin{aligned}\text{P}_{\mathbf{u}}^2&=(\text{uu}^{\text{T}})(\text{uu}^{\text{T}}) \\ &=\text{u}(\text{u}^{\text{T}}\text{u})\text{u}^{\text{T}}&\quad\text{(associative law)} \\ &=\text{uu}^{\text{T}}&\quad (\mathbf{u}\text{ is a unit vector)} \\ &=\text{P}_{\mathbf{u}}.\end{aligned} \nonumber \]