The discussion confirms the proof of the distributive property of vectors in Rn, specifically stating that for vectors u and v, and scalar c, the equation c(u+v) = cu + cv holds true. The proof involves distributing the scalar c across the components of the vectors, which are real numbers. Participants agree that the proof is straightforward and does not require complex steps, reinforcing the foundational nature of this property in vector algebra.
PREREQUISITESThis discussion is beneficial for students studying linear algebra, particularly those learning about vector properties and operations. It is also useful for educators seeking to clarify foundational concepts in vector mathematics.
B18 said:Homework Statement
Let u, and v be vectors in Rn, and let c be a scalar.
c(u+v)=cu+cv
The Attempt at a Solution
Proof:
Let u, v ERn, that is u=(ui)ni=1, and v=(vi)ni=1.
Therefore c(ui+vi)ni=1
At this point can I distribute the "c" into the parenthesis? For example:
=(cui+cvi)ni=1
=(cui)ni=1+(cvi)ni=1
=cu+cv.
B18 said:Ok I figured. So everything looks in order here? Just was expecting it to be a little more involved!