Suppose that x1, x2 ... xn are linearly independent and y is a vector which is not in the span of x1, x2... xn.

show that y, x1,x2...xn are also linearly independent?

Question

Suppose that x1, x2 ... xn are linearly independent and y is a vector which is not in the span of x1, x2... xn. 

show that y, x1,x2...xn are also linearly independent?

zzr0ck3r · Answer

Suppose it was not, then some vector in $\{x_i|i\in[n]\}\cup \{y\}$, call it $a$ is a combination of of other vectors in that same set. So $a$ is a lin combo of other vectors, if one of those vectors is $y$ solve for $y$ and you have that $y$ is a lin combo of the $x_i$. If $y$ is not in that lin combo, then the $x_i$ are not lin ind.

zzr0ck3r · Answer

contradiction....

zzr0ck3r · Answer

Maybe say something about if $a=y$ then $y$ is in the span of the $x_i$, so again a contradiction.

zzr0ck3r · Answer

does this make sense?

shawn · Answer

kinda though i'm still thinking though. I think you explained it the same way as this (see image) but in more a more compact way. https://postimg.org/image/ucncd2o1j/

zzr0ck3r · Answer

Let me know what parts confuse you, after you have sat with it for a bit. A little bit of confusion is the most opportune time to learn.

shawn · Answer

I was confused when you said If y is not in that lin combo, then the x_i are not lin ind. But now I got it. Thanks. :)