LinArg assignment 1

exercise

1. Lines in ${\mathbb{R}}^m$

a) Since $\mathbf{u}\in L$, $\mathbf{u}={\lambda }_2\mathbf{w}$ for some ${\lambda }_2\in \mathbb{R}$ $\mathbf{w}=\frac{\mathbf{u}}{{\lambda }_2}$ $L=\{{\lambda }_1\mathbf{w}:{\lambda }_1\in \mathbb{R}\}=\left\{\frac{{\lambda }_1}{{\lambda }_2}\mathbf{u}:{\lambda }_1\in \mathbb{R},{\lambda }_2\in \mathbb{R}\right\}=\{\lambda \mathbf{u}:\lambda \in \mathbb{R}\}$

b) $L_1=\{{\lambda }_1\mathbf{v}:{\lambda }_1\in \mathbb{R}\}$ $L_2=\{{\lambda }_2\mathbf{w}:{\lambda }_2\in \mathbb{R}\}$

$L_1=L_2$ ${\lambda }_1\mathbf{v}={\lambda }_2\mathbf{w}$ either ${\lambda }_1=0,{\lambda }_2=0$ meaning that $L_1\cap L_2=\{0\}$ or $\mathbf{v}=\lambda \mathbf{w},\lambda \in \mathbb{R}$ meaning that $\mathbf{w}$ is a element of $L_1$, and according to question a, they must be the same line.

c) $\mathbf{v}\cdot \mathbf{w}=0$ $\left(\begin{array}{c}{v}_1\\ v_2\end{array}\right)\left(\begin{array}{c}{d}_1\\ d_2\end{array}\right)=0$ $v_1{d}_1+v_2{d}_2=0$ $v_1=-\frac{d_2}{d_1}{v}_2$ $v_1=-\frac{d_2}{d_1}\lambda ,v_2=\lambda ,\lambda \in \mathbb{R}$ $\mathbf{v}=\left(\begin{array}{c}{v}_1\\ v_2\end{array}\right)=\lambda \left(\begin{array}{c}-\frac{d_2}{d_1}\\ 1\end{array}\right),\lambda \in \mathbb{R}$ $L=\{\lambda \left(\begin{array}{c}-\frac{d_2}{d_1}\\ 1\end{array}\right):\lambda \in \mathbb{R}\}$ and this is a line

2. Orthogonality and Linear independence

a) $\mathbf{v}\cdot \mathbf{w}=0$ $s+6+2s=0$ $s=-2$

b) $t=0$

c) $\mathbf{v}\cdot \mathbf{w}=0$ suppose $\mathbf{v},\mathbf{w}$ are linearly dependent: $\mathbf{w}=\lambda \mathbf{v}$ $\mathbf{v}\cdot \lambda \mathbf{v}=0$ $\lambda (\mathbf{v}\cdot \mathbf{v})=0$ $\mathbf{v}\cdot \mathbf{v}=0$ ${\sum }_{i=1}^n{v}_i^2=0$ There is no nonzero solution to this equation, so $\mathbf{v}$ and $\mathbf{w}$ must be linearly independent

3. Cauchy-Schwarz inequality

$$\sum _{i=1}^m{v}_i=\left(\begin{array}{c}1\\ 1\\ ⋮\\ 1\end{array}\right)\cdot \mathbf{v}\le \Vert 1\Vert \Vert \mathbf{v}\Vert =\Vert \mathbf{v}\Vert \le \sqrt{m}\Vert \mathbf{v}\Vert$$

$$\sum _{i=1}^m\sqrt{i}{v}_i=\left(\begin{array}{c}\sqrt{1}\\ \sqrt{2}\\ ⋮\\ \sqrt{m}\end{array}\right)\cdot \mathbf{v}\le \Vert \left(\begin{array}{c}\sqrt{1}\\ \sqrt{2}\\ ⋮\\ \sqrt{m}\end{array}\right)\Vert \Vert \mathbf{v}\Vert =\sqrt{1+2+\cdots +m}\Vert \mathbf{v}\Vert =\sqrt{\frac{m^2+m}{2}}\Vert \mathbf{v}\Vert \le m\Vert \mathbf{v}\Vert$$

Note that:

$$\sqrt{\frac{m^2+m}{2}}\le m⟺\frac{m^2+m}{2}\le m^2⟺m\le m^2\text{ for }m\ge 1$$

4. Linear Independence

a)

$${\mathbf{v}}_m=\sum _{j=1}^m(-1{)}^{j+1}{\mathbf{v}}_j=\left(\begin{array}{c}1+0+\cdots +0\\ 1+(-1)+0+\cdots +0\\ 0+(-1)+1+0+\cdots +0\\ ⋮\\ 0+\cdots +0+1\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\\ ⋮\\ 0\\ 1\end{array}\right)$$

b) ${\sum }_{i=1}^n{\lambda }_i{\mathbf{v}}_i=0$ Looking at the $i$-th coordinate gives ${\lambda }_i+{\lambda }_{i-1}=0$ (indices mod m) Hence ${\lambda }_i=(-1{)}^{i-1}{\lambda }_1$ ${\lambda }_1+{\lambda }_m=0⟹{\lambda }_1+(-1{)}^{m-1}{\lambda }_1=0$ Since m is odd then $(-1{)}^{m-1}=1$, so $2{\lambda }_1=0⟹{\lambda }_1=0$ and thus all ${\lambda }_i=0$, which is a trivial linear combination. Therefore ${\mathbf{v}}_1,\dots ,{\mathbf{v}}_m$ are linearly independent

5. Angle between vectors

$\cos (\alpha )=\frac{\mathbf{v}\cdot \mathbf{w}}{\Vert v\Vert \Vert w\Vert }=\frac{xz+xy+zy}{(\sqrt{x^2+y^2+z^2}{)}^2}=\frac{xz+xy+zy}{x^2+y^2+z^2}$ $x+y+z=0$ $(x+y+z{)}^2=0$ $x^2+y^2+z^2+2xy+2xz+2yz=0$ $xy+xz+yz=-\frac{1}{2}(x^2+y^2+z^2)$

$$\cos (\alpha )=-\frac{1}{2}\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=-\frac{1}{2}$$

6. Challenge 1.6

$$\{\begin{array}{l}{u}_1={\lambda }_1{v}_1+{\lambda }_2{w}_1\\ u_2={\lambda }_1{v}_2+{\lambda }_2{w}_2\end{array}$$

solve for ${\lambda }_1$: $\frac{w_2}{w_1}I-II$

$$\frac{w_2{u}_1}{w_1}-u_2=\frac{w_2{v}_1}{w_1}{\lambda }_1+w_2{\lambda }_2-{\lambda }_1{v}_2-{\lambda }_2{w}_2$$ $$⟹w_2{u}_1-u_2{w}_1=(w_2{v}_1-w_1{v}_2){\lambda }_1$$ $$⟹l_1=\frac{u_1{w}_2-u_2{w}_1}{v_1{w}_2-v_2{w}_1}$$

solve for ${\lambda }_2$: $\frac{v_1}{v_2}II-I$

$$\frac{u_2{v}_1}{v_2}-u_1=v_1{\lambda }_1+\frac{w_2{v}_1}{v_2}{\lambda }_2-{\lambda }_1{v}_1-{\lambda }_2{w}_1$$ $$⟹u_2{v}_1-u_1{v}_2=(w_2{v}_1-v_2{w}_1){\lambda }_2$$ $$⟹{\lambda }_2=\frac{u_2{v}_1-u_1{v}_2}{v_1{w}_2-v_2{w}_1}$$

Since $\mathbf{w}\ne \lambda \mathbf{v}$, meaning $v_1{w}_2-v_2{w}_1\ne 0$ , there will always be a solution of ${\lambda }_1$ and ${\lambda }_2$ q.e.d.