Projects space $\Rn{n}$ to some $\Rn{n-1}\subset \Rn{n}$ >[!important] All lines or planes you project with matrices *must* cross the origin. >[!info] Symmetry > $A^{\intercal}= A$ ### Determinant $\huge \det{A} = 0$ ### Inverse You cannot inverse a projection transformation. ### Using Vector Form. For lines in $\Rn2, \Rn3$ $\huge A= \frac{1}{\norms{\vec v}^{2}} \vec v {v}^{\intercal} = \hat v \hat v ^{\intercal} $ ### Using Normal Form For lines in $\Rn2$ or plane in $\Rn3$ that crosses the origin. $\huge A = I - \frac{1}{\norms{\vec n}^{2}}\vec n \vec n^\intercal$ $\huge\begin{align} A\mat{ 1\\1\\0} &= A\mat{0\\0\\1} \\ A\mat{ 1\\1\\0} - A\mat{0\\0\\1} &= \vec 0\\ A\mat{1\\1\\-1} &= \vec 0 \\ A &= I - \frac{1}{\norms{\vec n}^{2}}\vec n \vec n^\intercal \\ \end{align} $ $\huge\begin{align} \pa{I - \frac{1}{\norms{\vec n}^{2}}\vec n \vec n^\intercal }\mat{1\\1\\-1} &=\vec 0\\ \mat{1\\1\\-1} - \pa{ \frac{1}{\norms{\vec n}^2}\vec n \vec n^\intercal }\mat{1\\1\\-1} &= \vec 0 \\ \mat{1\\1\\-1} - \pa{ \hat n \hat n^\intercal }\mat{1\\1\\-1} &= \vec 0 \\ \hat n \hat n^\intercal \mat{1\\1\\-1} &= \mat{-1\\-1\\1} \\ \mat{x\\y\\z}\mat{x&y&z}=\mat{-1\\-1\\1} \end{align} $ $\huge \begin{align} \end{align}$