I Proving the Linear Transformation definition

[Aleoa]

HI .I'm trying to prove that, for a linear transformation, it is worth that:

f(a\bar{x}+b\bar{y})=af(\bar{x})+bf(\bar{y}) for every real numbers a and b.

Until now, I have proved by myself that

f(\bar{x}+\bar{y})=f(\bar{x})+f(\bar{y}).

and , using this result i proved that:

f(a\bar{v}) = f(\sum\bar{v})\text{(Sum it $a$ times)}.
=\sum f(\bar{v}).
=af(\bar{v}).
for every integer a.

How can I use this result in order to prove that
f(a\bar{v}) = af(\bar{v})
for every a\in \mathbb{R} ?

 

[PeroK]

The result you are trying to prove is the definition of a linear transformation. You can't prove this.

 

[Aleoa]

Hi @PeroK . Thanks for the reply.
I don't know, my book says it's possible to derive this result as i asked in my first post, even if it doesn't show how.

I upload the book's section I'm referring:

The book doens't start with the definition of Linear Trasformation, but try to deduce it.

 

[PeroK]

How is $f$ defined?

 

[Aleoa]

Simply says that f is a one-to-one affine trasformation that keeps the origin fixed.
Before, it says that the affine trasformation is a trasformation that carries lines to lines

 

[PeroK]

Simply says that f is a one-to-one affine trasformation that keeps the origin fixed.
Before, it says that the affine trasformation is a trasformation that carries lines to lines

Don't you think it might have been important to say that in the OP?

 

[Aleoa]

Don't you think it might have been important to say that in the OP?

I'm sorry. I'm here for improving :)

 

[Aleoa]

Anyone ?

 

[stevendaryl]

Well, in the section you quoted, it says "The proof of this fact is a little tricky, and we will prove it in an appendix..."

 

[Aleoa]

What i don't understand are the deductions done by my book, and you can see them in the 2 pictures i posted.

 

[Aleoa]

f(a\bar{v}) = f(\sum\bar{v})\text{(Sum it $a$ times)}.
=\sum f(\bar{v}).
=af(\bar{v}).
for every integer a.

One thing I am not able to understand is how use this demonstration in order to prove
f(a\bar{v}) =af(\bar{v}).
for every rational a.

My book does this implicitly in this image, but i don't understand what it did...

 

[stevendaryl]

One thing I am not able to understand is how use this demonstration in order to prove
f(a\bar{v}) =af(\bar{v}).
for every rational a.

My book does this implicitly in this image, but i don't understand what it did...

View attachment 222881

I'm not sure about the picture, either. But it seems to me that if it's true for integers, it has to be true for rationals, as well.

$f(n \cdot \frac{x}{n}) = n f(\frac{x}{n})$

Therefore, $f(\frac{x}{n}) = \frac{1}{n} f(x)$.

$f(\frac{m}{n} \cdot x) = m f(\frac{x}{n}) = \frac{m}{n} f(x)$

 

[Aleoa]

I'm not sure about the picture, either. But it seems to me that if it's true for integers, it has to be true for rationals, as well.

$f(n \cdot \frac{x}{n}) = n f(\frac{x}{n})$

Therefore, $f(\frac{x}{n}) = \frac{1}{n} f(x)$.

$f(\frac{m}{n} \cdot x) = m f(\frac{x}{n}) = \frac{m}{n} f(x)$

Thanks so much. Why can't i directly derive this proof to real values of a ?

 

[stevendaryl]

Thanks so much. Why can't i directly derive this proof to real values of a ?

Well, this is a pretty lame example, but suppose I define $f(x)$ this way:

• If $x$ is rational, then $f(x) = x$
• If $x$ is irrational, then $f(x) = 2x$

If $q$ is rational, then $f(qx) = q f(x)$. But if $q = \sqrt{2}$ and $x = \sqrt{2}$, then $f(qx) = f(2) = 2$, but $q f(x) = \sqrt{2} f(\sqrt{2}) = 4$

You need to know that $f(x)$ is continuous in order to prove $f(qx) = q f(x)$ when $q$ is irrational.