How is the x,y,z part a vector space?

[SpiffyEh]

Homework Statement

I uploaded the problem because its easier to see.

Homework Equations

The Attempt at a Solution

The solution is there from a practice exam. I don't understand the x,y,z part and how it fulfills the multiplication requirement of a vector space. Could someone please explain to me why it does?

 

[hgfalling]

In the solution they are handling the addition and multiplication at the same time.

They define u and v as vectors in the space, then show that u + dv (where d is any real) is also in the space.

So after multiplying and adding the vectors, you have a new vector:

\left(\begin{array}{cc} 4(a + da') + 3(b + db')\\0\\(c + dc') - 2\\(a + da') + (b + db')\end{array}\right)

So now all we have to do is show that all three (a+da'), (b+db'), (c+dc') are real numbers. Since a,b,c,d are all real, this is obviously the case, and hence the space is closed under addition and multiplication.

In the solution, they converted these expressions to (x,y,z), but that's unnecessary.

 

[SpiffyEh]

oh! That makes sense, i didn't see where they came from. Thank you