How can I prove this function is a linear function?

[tamtam402]

Homework Statement

I'm trying to prove f(r) = A $\bullet$ (r - kz) is a linear operator, but I don't understand why it is linear.

Homework Equations

A function of a vector is called linear if:
f(r1+r2) = f(r1) + f(r2) and f(ar) = af(r)

The Attempt at a Solution

f(r1+r2) = A $\bullet$ (r1+r2 - kz) = A$\bullet$r1 + A$\bullet$r2 - A$\bullet$kz

However, f(r1) + f(r2) = A$\bullet$r1 - A$\bullet$kz
+ A$\bullet$r2 - A$\bullet$kz

So, I just proved that the function is NOT a linear function, how great... :(

By the way, r = (x,y,z) in this book, if that changes anything.

 

[tamtam402]

Is it safe to assume this is an error in my book's manual? I even tried developing the vectors, here's my attempt. (Disregard the 3.7.1 part)

 

[tamtam402]

Ok, upon further investigation, the only way this can be linear is if I have to apply the -kz part BEFORE writing down my r vectors, which means

f(r1+r2) = (a1,a2,a3) $\bullet$ (x1+x2,y1+y2,0)

If that was the case, how in the hell was I supposed to know I had to do that (as opposed to what I did in my scanned solution)??!

 

[STEMucator]

I'm assuming by you saying r is a vector, you're saying r$\in$ℝⁿ. Notice that what you're trying to prove is f is closed under addition and scalar multiplication which implies f is linear.

Suppose that : r₁,r₂$\in$ℝⁿ

Then the way you would go about proving something like f(r₁+r₂)=f(r₁) + f(r₂) is to literally ADD the two vectors together since adding two objects together from the same field should result in something from the same field ( If it doesn't then you simply don't have anything to prove here ) and THEN you perform the operator f on the NEW object. Simply separating the newly formed object into its two founding objects is a simple task at this point which should result in your answer.

 

[Chestermiller]

Homework Statement

I'm trying to prove f(r) = A $\bullet$ (r - kz) is a linear operator, but I don't understand why it is linear.

Homework Equations

A function of a vector is called linear if:
f(r1+r2) = f(r1) + f(r2) and f(ar) = af(r)

The Attempt at a Solution

f(r1+r2) = A $\bullet$ (r1+r2 - kz) = A$\bullet$r1 + A$\bullet$r2 - A$\bullet$kz

However, f(r1) + f(r2) = A$\bullet$r1 - A$\bullet$kz
+ A$\bullet$r2 - A$\bullet$kz

So, I just proved that the function is NOT a linear function, how great... :(

By the way, r = (x,y,z) in this book, if that changes anything.

k is the unit vector in the z direction, and r is given by:

r = x i + y j + z k

Therefore, f (r) = A $\bullet$ (x i + y j)

Alternately,

f(r1) = A$\bullet$ (r1 - k z1)

f(r2) = A$\bullet$ (r2 - k z2)

Try it now.

 

[Ray Vickson]

Homework Statement

I'm trying to prove f(r) = A $\bullet$ (r - kz) is a linear operator, but I don't understand why it is linear.

Homework Equations

A function of a vector is called linear if:
f(r1+r2) = f(r1) + f(r2) and f(ar) = af(r)

The Attempt at a Solution

f(r1+r2) = A $\bullet$ (r1+r2 - kz) = A$\bullet$r1 + A$\bullet$r2 - A$\bullet$kz

However, f(r1) + f(r2) = A$\bullet$r1 - A$\bullet$kz
+ A$\bullet$r2 - A$\bullet$kz

So, I just proved that the function is NOT a linear function, how great... :(

By the way, r = (x,y,z) in this book, if that changes anything.

Note: $z$ is the third component of $\textbf{r}$, so if you have $\textbf{r}_1$ and $\textbf{r}_2$ you need $z_1$ and $z_2$.

RGV

 

[tamtam402]

Yeah thanks everyone. I was assuming the kz wasn't the "same z" as the z component from the r vectors, which means I considered r1-z as (x1,y1,z1)-(0,0,z) = (x1,y1,z1-z).