I have a question about proving that a set is not a vector space. In my Linear Algebra textbook, we have this example:

"Let V be the set of all ordered pairs of real numbers, with the standard operation of addition and the nonstandard definition of scalar multiplication listed below."

$c( x_1 , x_2) = ( cx_1, 0)$

The book says that the first 9 axioms pass, but that the 10th axiom (Multiplicative Identity Property) fails because:

$1(1,1) = (1,1) \neq (1,0)$

But I am not certain as to why... Is it because $1(x_1, x_2) = (1x_1,1x_2)$ which simplifies to $(x_1,x_2)$, thus resulting in $(1,1)$ as opposed to $(1,0)$?

Though, I have doubts, because at the same time I can see 1 being c in the way scalar multiplication is defined, and thus $1(x_1, x_2) = (x_1, 0)$.


1 Answer 1


According to the axiom that we mentioned, you should have $1(1,1)=(1,1)$. But what we actually have for this scalar product is $1(1,1)=(1,0)$. Since $(1,1)\neq(1,0)$, that axiom doesn't hold and therefore we don't have a vector space.

  • $\begingroup$ Just for clarification then: When proving the 10th axiom, I should first apply the multiplicative identity with standard scalar multiplication and see how it compares to the nonstandard scalar multiplication? $\endgroup$
    – hiyumi
    Apr 28, 2019 at 10:43
  • $\begingroup$ No! You should apply the nonstandard scalar multiplication and then see whether or not you got the vector that you started with. $\endgroup$ Apr 28, 2019 at 11:30

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