math121b: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.

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.

math121b/about_vectors_and_tensors.txt · Last modified: 2020/02/10 13:25 by pzhou