Thread: Linear algebra vector space View Single Post
 P: 351 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.