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\text { 1) Row space } \begin{aligned}& \operatorname{span}\{(3),(-6)\} \\& =\operatorname{span}\{(1)\}\}\end{aligned}nomalizing (1) (w) we get \left(\begin{array}{c}1 / \sqrt{2} \\ 1 / \sqrt{2}\end{array}\right) \sin \varphi, \sqrt{1+1}=\sqrt{2}2) Col spa e span \{((3)\}. length of vector is \sqrt{9+36}-3 \sqrt{5} so nomali - zing column vector \Rightarrow(y / \sqrt{5},-2 / \sqrt{5})=4,\begin{array}{l}\omega=\left[\begin{array}{l}2 / \sqrt{5} \\ 5\end{array}\right]-u_{2} \\\end{array}\begin{array}{rl}M & U \Sigma V^{T} \\& =\left[\begin{array}{ll}1 / \sqrt{5} & 2 / \sqrt{5} \\-2 / \sqrt{5} & y / \sqrt{5}\end{array}\right]\left[\begin{array}{ll}x & 0 \\0 & y\end{array}\right]\left[\begin{array}{ll}y / \sqrt{2} & 1 / \sqrt{2} \\-1 / \sqrt{2} & 1 / \sqrt{2}\end{array}\right]\end{array}As 2 a n k(M)=1, \quad \Rightarrow one singular value,y=0.\begin{array}{l}{\left[\begin{array}{c}3.3 \cdot \\-6-6\end{array}\right]=\left[\begin{array}{cc}x / \sqrt{10} & x / \sqrt{10} \\-2 \cdot x / \sqrt{10} & -2 x \cdot 10\end{array}\right]} \\\Rightarrow x=3 \sqrt{10} \\\end{array}Singular value \quad 3 \sqrt{10}, 0 \Rightarrow \sqrt{90}, \ldots ...