# How do I prove a component of a vector is 0 for all i in I?

Let $E_1 = (1, 0, \ldots ,0)$, $E_2 = (0, 1, 0, \ldots ,0)$, ... , $E_n = (0, ... ,0, 1)$ be the standard unit vectors of $R^n$. Let $x_1, \ldots ,x_n$ be numbers. What is $x_1E_1 + \ldots + x_nE_n$? Show that if $x_1E_1 + \ldots + x_nE_n = 0$, then $x_i = 0$ for all $i$.

My attempt:

$x_1E_1 = (x_1, 0, …, 0)$

$x_nE_n = (0, 0, …, x_n)$

$x_1E_1 + … + x_nE_n = (x_1, 0, …, 0) + … + (0, 0, …, x_n)$

= $(x_1, …, x_n) = (0, …, 0)$

Let $X = ( x_1, …, x_n)$

Here every $x_i$ in $X$ corresponds to $0$ in vector $0$. So, $x_i = 0$ for all $i$.

Please, check and see if that makes sense. Thanks.

It makes perfect sense. I would say it was right :)

The only thing I might do just to be as clear as possible, is to write the equations as an array:

$x_1=0$

$x_2=0$

...

$x_n=0$

But that might just be overkill.

• I hear it's frowned upon here to just post one's appreciation, but I thank you. – IndianOcean Apr 1 '14 at 19:43
• This is a place for helping and learning, there was nothing to teach as you got it right already, I just gave you the confirmation you were seeking. – Ellya Apr 1 '14 at 19:47