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 The whole point of doing abstract vector space, is to show that, we can do linear algebra without using the crutches of 'basis'.
- Let $V = \R^2$. Can you give a basis of linear functions on $V$?
+. Let $V = \R^2$. Can you give a basis of linear functions on $V$?
- Suppose $V = \R^2$ with standard coordinates $(x,y)$ on $\R^2$. We choose a new basis $e_1 = (1,1), e_2=(0,2)$. Now, we are going to find the dual basis for $e_1, e_2$. These are two elements $h^1, h^2$ in $V^*$, or equivalently, we have two functions $h^1(x,y)$ and $h^2(x,y)$ on $\R^2$ (these superscripts $1,2$ are just labels, don't confuse them with power of a function). I will give you a hint of how to find $h^1(x,y)$. First of all, $h^1(x,y)$ is a linear function, so it is of the form $h^1(x,y) = ax+by$ for some unknown constants $a,b$. Then, we recall that the function $h^1$ evaluate on the point $e_1$ equals to $1$. Where is $e_1$? Recall $e_1=(1,1)$, so we plug in the coordinate $e_1$ as $(x,y)$ into $h^1(x,y)$, and we require $h^1(1,1)=1$. Similarly, we also require that the function $h^1$'s value on $e_2$ equal to $0$. This will help you determine $h^1$. Try on your own determine $h^2$. Draw on a piece of paper, the lines of $h^1(x,y)=-1, 0, 1$ and $h^2(x,y) = -1,0,1$. You should see a skewed a grid.
+. Suppose $V = \R^2$ with standard coordinates $(x,y)$ on $\R^2$. We choose a new basis $e_1 = (1,1), e_2=(0,2)$. Now, we are going to find the dual basis for $e_1, e_2$. These are two elements $h^1, h^2$ in $V^*$, or equivalently, we have two functions $h^1(x,y)$ and $h^2(x,y)$ on $\R^2$ (these superscripts $1,2$ are just labels, don't confuse them with power of a function). I will give you a hint of how to find $h^1(x,y)$. First of all, $h^1(x,y)$ is a linear function, so it is of the form $h^1(x,y) = ax+by$ for some unknown constants $a,b$. Then, we recall that the function $h^1$ evaluate on the point $e_1$ equals to $1$. Where is $e_1$? Recall $e_1=(1,1)$, so we plug in the coordinate $e_1$ as $(x,y)$ into $h^1(x,y)$, and we require $h^1(1,1)=1$. Similarly, we also require that the function $h^1$'s value on $e_2$ equal to $0$. This will help you determine $h^1$. Try on your own determine $h^2$. Draw on a piece of paper, the lines of $h^1(x,y)=-1, 0, 1$ and $h^2(x,y) = -1,0,1$. You should see a skewed a grid.
- (Quotient vector space). Let $V$ be a vector space, $W \subset V$ be a subspace. We define a set $V / W$ as following,
-  - two elements $v, v'$ in $V$ are called equivalent, if $v - v' \in W$. It is denoted as $v \sim v'$.
-  - an equivalence class of $V$ is a maximal subset of $V$ where every two elements in it are equivalent.
-    - let $v \in V$, there is exactly one equivalence class that contains $v$, and is denoted as $[v]$.
-  - $V / W$ is the set of equivalence classes of $V$, that means, an element in $V/W$ is an equivalence class of $V$.
-For example, let $V = \R^2$, and $W$ is the one-dimensional subspace defined by $x=y$, or $W = span_\R (1,1)$. Then $(2,3)$ and $(4,5)$ are in the same equivalence classes, because $(2,3)-(4,5)=(-2,-2)$ and is  in $W$.
-Show that the set $V / W$ can be equipped with a vector space structure

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