The problem we will be looking at is in Chapter VS, M11.
Let V be the set in C2 with the usual vector addition, but with scalar multiplication defined by: \[ \alpha \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} \alpha y \\ \alpha x\end{bmatrix} \]
Determine whether or not V is a vector space with these operations.
Before we proceed with this proof, we need to recall the axioms of a vector space. I really like this particular posting of the axioms. They are easy to follow. http://www.math.uconn.edu/~glaz/math2210f16/Section%20Handouts/sec4_1.pdf
The axioms are as follows:
A vector space is a nonempty set V of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real numbers), subject to the ten axioms below. The axioms must hold for all u, v and w in V and for all scalars c and d.
## [1] "1. u . v is in V"## [1] "2. u + v = v + u"## [1] "3. (u + v)+ w = u + (v + w)"## [1] "4. There is a vector (called the zero vector) 0 in V such that u + 0 = u"## [1] "5. For each u in V, there is vector -u in V satisfying u + (-u) = 0"## [1] "6. cu is in V"## [1] "7. c(u + v) = cu + cv"## [1] "8. (c + d)u = cu + du"## [1] "9. (cd)u =c(du)"## [1] "10. 1u = u"At it’s core, we want to show that the axioms hold true for our problem. If they hold true, then we can declare that V is indeed a vector space. On the other hand, if we can find a counter example that goes against the axioms, then we can conclude that set V is NOT a vector space. Only one of the axioms needs to fail.
Proof: The way vector V is defined, we can easily find a counter example. Our set V looks a lot like axiom 10. You have some scalar multiplied against u. Lets set our scalar to 1 and choose x=1 and y=2.
\[ \alpha \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} \alpha y \\ \alpha x \end{bmatrix}\\ \\ 1\begin{bmatrix} 1 \\ 2 \end{bmatrix}\neq \begin{bmatrix} 1\cdot 2 \\ 1\cdot 1 \end{bmatrix}=\begin{bmatrix} 2 \\ 1 \end{bmatrix}\\ \\ \begin{bmatrix} 1 \\ 2 \end{bmatrix}\neq \begin{bmatrix} 2 \\ 1 \end{bmatrix} \]
By contradiction, we have shown that set V contradicts axiom 10, therefore it is NOT a vector space.