# Linear Algebra - Show that $V$ is not a vector space

Let $V = \{(x,y,z) | x, y, z \in R\}$. Define addition and scalar multiplication on $V$ as follows:

$$(x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1,y_1+y_2,z_1+z_2)$$ $$c(x_1,y_1) = (2cx_1,cy_1)$$

where $c$ is any real number.

Show that $V$, with respect to these operations of addition and scalar multiplication, is not a vector space by showing that one of the vector space axioms does not hold.

Since I'm new to Linear Algebra, I don't understand how to approach this question, any help would be much appreciated!

• You have to check that at least one of the axioms listet here does not hold: en.wikipedia.org/wiki/Vector_space – b00n heT Feb 12 '14 at 12:05
• Presumably, somewhere you have been shown a list of about 10 vector space axioms. The question asks you to go through that list, one axiom at a time, until you find one that doesn't work for the set $V$. – Gerry Myerson Feb 12 '14 at 12:05
• I would definitely go with the existance of the identity element of scalar multiplication (iow: set $c=1$) – b00n heT Feb 12 '14 at 12:07
• Your second equation only uses ordered pairs, but the space contains orders triples. – MPW Feb 12 '14 at 12:20

Hint: One of the axioms is that $av + bv = (a+b)v$ for scalars $a,b$ and vector $v$. See how that works with your scalar multiplication.
HINT: You need commutativity. Notice that the first coordinate of the sum is $x_1$. What if you added them in a different order $$(x_2,y_2,z_2)+(x_1,y_1,z_1)?$$
Under addition, $V$ should be an abelian group. Especially there should be a neutral element $(x_1,y_1,z_1)$ such that $(x_1,y_1,z_1)+(x_2,y_2,z_2)= (x_2,y_2,z_2)$ for all $(x_2,y_2,z_2)$.
Also, scalar multiplication with $c=1$ should be the identity. Not to mention that it should be defeined on elements of $V$ in the first place (your definition works with pairs, not triples).