Question

( x ) | x in R
( 1 )

is it subspace in R^2

Answer 1

let's do scalar multiple

c* (x) = cx
(1) c

test for linearity

no because we need and 1 and we get a c

Answer 2

@zarkon, can you confirm

Answer 3

is that the vactor \[\left[\begin{matrix}x \\ 1\end{matrix}\right]\]

Answer 4

vector

Answer 5

yes

Answer 6

subspace contains the zero vector

Answer 7

your first post also works

Answer 8

how does zero vector explanation work

Answer 9

the zero vector is \[\left[\begin{matrix}0\\ 0\end{matrix}\right]\]

\[\left[\begin{matrix}x \\ 1\end{matrix}\right]\] is not of that form

Answer 10

oh , since 1 can't ever be 0

Answer 11

you could also do this

\[\left[\begin{matrix}x \\ 1\end{matrix}\right]+\left[\begin{matrix}y \\ 1\end{matrix}\right]=\left[\begin{matrix}x+y \\ 2\end{matrix}\right]\]
which is also not of the original form

Answer 12

correct

Answer 13

nice, thanks