Linear Algebra Tutorial Class, Week 20

Question One

Question

Let $\{{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3\}$ be the standard vectors in ${\mathbb{R}}^3$. Determine whether

$$\text{span}({\mathbf{e}}_1,-2{\mathbf{e}}_2,{\mathbf{e}}_1+{\mathbf{e}}_2-{\mathbf{e}}_3)={\mathbb{R}}^3$$

Solution

If the three vectors span ${\mathbb{R}}^3$, then they must be able to generate any general vector in the plane $\mathbf{x}=[x_1,x_2,x_3]\in {\mathbb{R}}^3$, such that if we let $c_1,c_2,c_3\in \mathbb{R}$ and $\mathbf{w}\in {\mathbb{R}}^3$ satisfy the following equation:

$$c_1({\mathbf{e}}_1)+c_2(-2{\mathbf{e}}_2)+c_3({\mathbf{e}}_1+{\mathbf{e}}_2-{\mathbf{e}}_3)=\mathbf{w}$$ $$⟺(c_1+c_3){\mathbf{e}}_1+(-2c_2+c_3){\mathbf{e}}_2+(-c_3){\mathbf{e}}_3=\mathbf{w}$$ $$⟺\left[\begin{array}{c}{c}_1+c_3\\ -2c_2+c_3\\ -c_3\end{array}\right]=\mathbf{w}=\mathbf{x}$$ $$⟺\{\begin{array}{l}{c}_1+c_3=x_1\\ -2c_2+c_3=x_2\\ -c_3=x_3\end{array}$$ $$⟺c_3=-x_3,c_2=-\frac{x_2+x_3}{2},c_1=x_1+x_3$$

Hence, the vectors span ${\mathbb{R}}^3$.$\square$

Question Two

Question

Find the span of the following vectors:

$$\mathbf{u}=\left[\begin{array}{c}0\\ -2\\ 0\\ 5\end{array}\right],\mathbf{v}=\left[\begin{array}{c}3\\ 1\\ 1\\ 0\end{array}\right]$$

and

$$\mathbf{u}=\left[\begin{array}{c}1\\ -1\\ 0\\ 2\end{array}\right],\mathbf{v}=\left[\begin{array}{c}0\\ -\frac{1}{2}\\ \frac{3}{2}\\ 1\end{array}\right],\mathbf{w}=\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right]$$

Solution

$$\begin{array}{rl}{\text{span}}_1& =c_1\mathbf{u}+c_2\mathbf{v}\\ & =c_1\left[\begin{array}{c}0\\ -2\\ 0\\ 5\end{array}\right]+c_2\left[\begin{array}{c}3\\ 1\\ 1\\ 0\end{array}\right]\\ & =\left[\begin{array}{c}{c}_2\\ -2c_1+c_2\\ c_2\\ 5c_1\end{array}\right]\end{array}$$

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Question Three

Question

Find a value for $c$ so that the following vectors are linearly independent:

$$\mathbf{u}=\left[\begin{array}{c}1\\ 1-c\end{array}\right],\mathbf{v}=\left[\begin{array}{c}1\\ 0\end{array}\right]$$

Solution

By inspection, we can see that the solution is such that:

$$c\ne 1$$

Formerly however, if we let $c_1,c_2\in \mathbb{R}$, then the vectors must satisfy the equation $c_1\mathbf{u}+c_2\mathbf{v}=0⟹c_1=c_2=0$.

$$\left[\begin{array}{c}{c}_1+c_2\\ c_1(1-c)\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]⟺\{\begin{array}{l}{c}_1+c_2=0\\ c_1(1-c)=0\end{array}:c=1⟹c\ne 1$$

if linearly independent (since if $c=1$ then the vectors are linearly dependent, but if $c=k:k\ne 1$ then the vectors are linearly independent).

Question Four

Question

Find the dot product, the length, and the angle between the vectors:

$$\mathbf{u}=\left[\begin{array}{c}0\\ \frac{1}{2}\\ -\frac{1}{2}\end{array}\right],\mathbf{v}=\left[\begin{array}{c}\frac{1}{2}\\ \frac{2}{3}\\ \frac{2}{3}\end{array}\right]$$

Then, verify the Cauchy-Schwarz inequality and the triangle inequality.

Solution

$$\begin{array}{rl}\mathbf{u}\cdot \mathbf{v}& =\left[\begin{array}{c}0\\ \frac{1}{2}\\ -\frac{1}{2}\end{array}\right]\cdot \left[\begin{array}{c}\frac{1}{2}\\ \frac{2}{3}\\ \frac{2}{3}\end{array}\right]\\ & =0+\frac{2}{6}-\frac{2}{6}\\ & =0\therefore \text{orthogonal}\end{array}$$

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