# Vectors in index notation

1. Feb 2, 2013

### redstone

given the vector in the first equation below, does that necessarily imply the third equation, as shown?
$${{u}_{a}}{{e}^{a}}={{x}_{a}}{{e}^{a}}$$
$${{u}_{a}}{{e}^{l}}g_{l}^{a}={{x}_{a}}{{e}^{l}}g_{l}^{a}$$
$${{u}_{a}}{{e}^{l}}={{x}_{a}}{{e}^{l}}$$

2. Feb 2, 2013

### oli4

Hi redstone
No
In the first line, you have a single scalar equation, what you have written is that the dot product of u with e is the same as the dot product of x with e.
That of course does not mean that u = x unless your space is of dimension 1
In the third line, you have many equations saying that all components of u and x for any given index give the same result when multiplied by any component (but the same for u and x) of e, which lets you conclude that u = x

3. Feb 2, 2013

### vela

Staff Emeritus
That's not true. What if $e^l=0$ for all $l$?

Regarding the original post, it looks like you're trying to cancel $g^a_l$ from both sides of the equation. You can't do that.

4. Feb 2, 2013

### oli4

Hi vela
well, of course if e is the null vector there isn't much to conclude neither in the first equation nor the third.
I wasn't trying to tell some general always valid truth, I was just trying to show how the first equation couldn't possibly lead to the third without just saying that, indeed, you can't do the second step :)

5. Feb 3, 2013

### CompuChip

The choice of e as opposed to any other letter suggests that $e^l$ is not an arbitrary vector but a basis of the vector space. Although vela has a point that we can't really assume that until redstone gives us a bit more information on what he is actually looking for.