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# System prepended metadata

title: Diff Geometry - Activities
tags: [teaching]

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title: Diff Geometry - Activities
tags: teaching
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# Review
## Vector spaces and inner products

* Below $A$ and $B$ are defined. They are subsets of $\mathbb R^2$. Using the usual addition and scalar multiplication in $\mathbb R^2$, they are _not_ vector spaces. Explain why. $$A = \{(x,y)\in\mathbb R^2\ :\ x=0\ \ \text{or}\ \ y=0\}$$ $$B = \{(x,y)\in\mathbb R^2\ :\ y \le x\ \ \text{and}\ \ y\ge-x\}$$

* Is the set $\{(0,0)\}$ a vector space? (using usual addition and scalar multiplication operations) What about $\{(1,0)\}$? Explain.

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Fix two vectors ${\bf v_1} = \begin{bmatrix}1 \\0 \\1\end{bmatrix}$ and ${\bf v_2} = \begin{bmatrix}2 \\1 \\1\end{bmatrix}$ in $\mathbb R^3$. Define an inner product on $\mathbb R^2$ as follows: if ${\bf a}=(a^1,a^2)$ and ${\bf b}=(b^1,b^2)$ are vectors in $\mathbb R^2$, define $$\langle{\bf a}, {\bf b}\rangle = \left(a^1{\bf v_1}+a^2{\bf v_2}\right)\cdot\left(b^1{\bf v_1}+b^2{\bf v_2}\right).$$ (Here $\cdot$ is the usual dot product.) 
With this definition of inner product, 
* compute $\langle(1,1), (0,-1)\rangle$; and,
* find a vector $(b^1,b^2)$ that is orthogonal, for this inner product, to $(1,0)$.

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Define ${\bf v_1} = (1,1)$ and ${\bf v_2} = (0,1)$. Suppose that $L:\mathbb R^2\to\mathbb R^2$ is a linear transformation, and you know that $L({\bf v_1}) = (1,-2)$ and $L({\bf v_2}) = (0,2)$. 
* Can you determine what $L((-1,0))$ must be? 
* Is the entire linear transformation determined by knowing $L({\bf v_1})$ and $L({\bf v_2})$? Why or why not?

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Let ${\bf a} = (0,2,1)$, ${\bf b} = (3,-1,-2)$, and ${\bf c} = (1,1,1)$. Write an equation for _the plane through_ ${\bf c}$ _and perpendicular to_ ${\bf b}$. Is the origin $(0,0,0)$ a point that is contained in that plane?

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Let ${\bf \alpha}(t) = \begin{bmatrix}1+t \\1+t^2 \\1-t-t^2\end{bmatrix} = (1+t){\bf e_1} + (1+t^2){\bf e_2} + (1-t-t^2){\bf e_3}$.

Given the following basis of $\mathbb R^3$, $\{{\bf b_1}, {\bf b_2}, {\bf b_3}\}$ where $${\bf b_1} = \begin{bmatrix}1 \\1 \\0\end{bmatrix}, {\bf b_2} = \begin{bmatrix}1 \\-1 \\0\end{bmatrix}, {\bf b_3} = \begin{bmatrix}1 \\1 \\1\end{bmatrix},$$ find the component functions of ${\bf \alpha}$ in the basis $\{{\bf b_1}, {\bf b_2}, {\bf b_3}\}$.

