Example problem of [[../02 Areas/Math/Singular Value Decomposition|Singular Value Decomposition]]. $\large \begin{align} A &= \mat{ 1.1&-0.2&1.5\\ -2.3&-1.4&0.5 }\\ &= \underbrace{U}_{{2\times 2}} \underbrace{\Sigma}_{{2\times 3}} \underbrace{V^{\intercal}}_{3\times 3} \end{align} $ $ \large \begin{align} G =A^\intercal A &= \mat{ \frac{3}{2} & 3 & \frac{1}{2}\\ 3&2&-1 \\ \frac{1}{2} & -1 & \frac{5}{2} } \\ \det(G-tI) &= -t^3 + 11t^2 - 24t\\ &\therefore \begin{matrix} \lambda_{1} = 6 & \lambda_{2}=3 & \lambda_{3}=0 \end{matrix} \end{align} $ $\large \begin{align} E_{8} &= \op{Nul}(G-8I) \\ &\sim \begin{cases} x_{1}-2x_{2}&=0\\ x_{3}&= 0 \\ 0 &= 0 \\ \end{cases}\\ E_{8} &= \op{span}\left\{ \mat{2\\1\\0} \right\} \\ \\ E_{3} &= \op{Nul}(G-3I) \\ &= \op{span}\left\{ \mat{ \frac{1}{5} \\ -\frac{2}{5} \\ 1 } \right\} \\ \\ E_{0} &= \op{Nul}(G-0I) \\ &= \op{span}\left\{ \mat{ -1\\2\\1 } \right\} \end{align} $ $\huge\begin{align*} \hat v_{1} &= \mat{ \frac{2}{\sqrt{ 5 }} \\ \frac{1}{\sqrt{ 5 }}\\ 0 }\\ \hat v_{2} &= \mat{ \frac{1}{\sqrt{ 30 }} \\ \frac{-2}{\sqrt{ 30 }} \\ \frac{5}{\sqrt{ 30 }} \\ }\\ \hat v_{2} &= \mat{ \frac{-1}{\sqrt{ 6 }} \\ \frac{2}{\sqrt{ 6 }} \\ \frac{1}{\sqrt{ 6 }} \\ }\\ \end{align*} $ $\huge V = \mat{ \frac{2}{\sqrt{ 5 }} & \frac{1}{\sqrt{ 30 }} & \frac{-1}{\sqrt{ 6 }} \\ \frac{1}{\sqrt{ 5 }}& \frac{-2}{\sqrt{ 30 }} & \frac{2}{\sqrt{ 6 }} \\ 0 & \frac{5}{\sqrt{ 30 }} &\frac{1}{\sqrt{ 6 }} \\ } $ $\huge \begin{align} \sigma_{1} &= \sqrt{ \lambda_{1} } \\ \sigma_{2} &= \sqrt{ \lambda_{2} } \\ \Sigma &= \mat{ \sqrt{ 8 } & 0 & 0 \\ 0 & \sqrt{ 3 } &0 } \end{align} $ $\large \begin{align*} \frac{1}{\sigma_{1}} A \hat v_{1} &= \hat u_{1} = \mat{ \frac{1}{\sqrt{ 10 }}\\ -\frac{3}{\sqrt{ 10 }} } \\ \frac{1}{\sigma_{2}} A \hat v_{2} &= \hat u_{2} = \mat{ \frac{3}{\sqrt{ 10 }} \\ \frac{1}{\sqrt{ 10 }} } \\ U &= \mat{ \frac{1}{\sqrt{ 10 }} & \frac{3}{\sqrt{ 10 }}\\ -\frac{3}{\sqrt{ 10 }} & \frac{1}{\sqrt{ 10 }} } \end{align*} $ $\large A = \mat{ \frac{1}{\sqrt{ 10 }} & \frac{3}{\sqrt{ 10 }}\\ -\frac{3}{\sqrt{ 10 }} & \frac{1}{\sqrt{ 10 }} }\mat{ \sqrt{ 8 } & 0 & 0 \\ 0 & \sqrt{ 3 } &0 }\mat{ \frac{2}{\sqrt{ 5 }} & \frac{1}{\sqrt{ 30 }} & \frac{-1}{\sqrt{ 6 }} \\ \frac{1}{\sqrt{ 5 }}& \frac{-2}{\sqrt{ 30 }} & \frac{2}{\sqrt{ 6 }} \\ 0 & \frac{5}{\sqrt{ 30 }} &\frac{1}{\sqrt{ 6 }} \\ } $ $\frac{1}{\sqrt{ 5 }}\mat{2\\1\\0}$ is the [[Unit Vector]] in $\R^3$ such that $\lvert \lvert A\hat v_{1} \rvert \rvert = \sqrt{ 8 }$ is maximal, $\hat u_{1} \parallel A \hat{v}_{1}$ $\frac{1}{\sqrt{ 30 }}\mat{1\\-1\\5}$ is the [[Unit Vector]] in $\R^3$ such that $\lvert \lvert A\hat v_{2} \rvert \rvert = \sqrt{ 3 }$ is maximal, $\hat u_{2} \parallel A \hat{v}_{2}$ $\frac{1}{\sqrt{ 6 }}\mat{-1\\2\\1}$ is the [[Unit Vector]] in $\R^3$ such that $\lvert \lvert A\hat v_{3} \rvert \rvert = 0$ is minimal [[../02 Areas/Math/Singular Value Decomposition Truncation|Singular Value Decomposition Truncation]]: $\op{rank}(A) = 2$, find the best fitting [[../02 Areas/Math/Rank|Rank]] $1$ [[../02 Areas/Math/Matrix|Matrix]] to $A$. $ \begin{align} A^\dagger &= U \mat{ \sqrt{ 8 } &0&0 \\0&0&0 }V^\intercal \\ &= \mat{ \frac{1}{\sqrt{ 10 }} & \frac{3}{\sqrt{ 10 }}\\ -\frac{3}{\sqrt{ 10 }} & \frac{1}{\sqrt{ 10 }} }\mat{ \sqrt{ 8 } & 0 & 0 \\ 0 & 0 &0 }\mat{ \frac{2}{\sqrt{ 5 }} & \frac{1}{\sqrt{ 30 }} & \frac{-1}{\sqrt{ 6 }} \\ \frac{1}{\sqrt{ 5 }}& \frac{-2}{\sqrt{ 30 }} & \frac{2}{\sqrt{ 6 }} \\ 0 & \frac{5}{\sqrt{ 30 }} &\frac{1}{\sqrt{ 6 }} \\ } \\ &= \mat{ 0.8 & 0.4 & 0 \\ -2.4 & -1.2 & 0 } \end{align} $