# Exercises about Vectors and Tensors

If you are unfamiliar with the sentence “elements of a set” or “elements of a vector space”, you can take a look at Sets.

## Getting Familiar with basis free notation

The whole point of doing abstract vector space, is to show that, we can do linear algebra without using the crutches of 'basis'.

1. Let $V = \R^2$. Can you give a basis of linear functions on $V$?

2. 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.