LINEAR ALGEBRA: Show that |a x| = |a| |x| for all real numbers a

[VinnyCee]

How does one prove this statement?

I have no idea how to start. Can someone help?

Maybe it has something to do with that Cauchy-Schwartz inequality?

 

[VinnyCee]

I guess using bilinearity of dot product spaces will work, right?

Since (a x) * y = a (x * y) = x * (ay)

 

[quasar987]

I don't think bilinearity of dot product space is something someone struggling to prove |a x| = |a| |x| knows about!

You could use the definition of |.|, namely that

$$|ax|=\left\{ \begin{array} {c} -ax \ \ \mbox{for} \ \ ax<0 & +ax \ \ \mbox{for} \ \ ax\geq 0 \end{array}$$

and treat the three cases a<0, a>0, a=0 separetely.

 

[VinnyCee]

$$|a|=\left\{ \begin{array} {c} -a \ \ \mbox{for} \ \ a<0 & +a \ \ \mbox{for} \ \ a\geq 0 \end{array}$$

$$|x|=\left\{ \begin{array} {c} -x \ \ \mbox{for} \ \ x<0 & +x \ \ \mbox{for} \ \ x\geq 0 \end{array}$$

Does this somehow "prove" that |a x| = |a| |x| $\in$ R?

If not, what would count as "proof"?

 

[quasar987]

I don't understand your question. My post was meant as a hint to the OP.

 

[geoffjb]

I don't understand your question. My post was meant as a hint to the OP.

I think he means, is what you posted sufficient to constitute a proof?

 

[Office_Shredder]

EDIT: Wrong absolute value. I should read the thread

 

[HallsofIvy]

I don't think bilinearity of dot product space is something someone struggling to prove |a x| = |a| |x| knows about!

You could use the definition of |.|, namely that

$$|ax|=\left\{ \begin{array} {c} -ax \ \ \mbox{for} \ \ ax<0 & +ax \ \ \mbox{for} \ \ ax\geq 0 \end{array}$$

and treat the three cases a<0, a>0, a=0 separetely.

If you know about "bilinearity of dot product" then surely you understood that the question was about the lengths of vectors, not about absolute value? "|ax|= -ax for ax< 0" makes no sense because the left side of the equation is a number and the right side is a vector. Also, there is no order defined on a vector space.

 

[Galileo]

Isn't |ax|=|a||x| by definition of a norm on a vector space?

 

[radou]

Let a be a real number, and $$\vec{v} \in V^2(O)$$, for simplicity.

$$\left|a\right|\cdot\left|\vec{v}\right|=\left|a\right|\sqrt{v_{x}^2+v_{y}^2}=\sqrt{\left|a\right|^2(v_{x}^2+v_{y}^2)}=\sqrt{a^2(v_{x}^2+v_{y}^2)}=\sqrt{(av_{x})^2+(av_{y})^2}=\left|a \cdot \vec{v}\right|$$. Could this be considered as a proof?

 

[quasar987]

I didn't realize that post #2 was by the OP himself, lol. And for some reason I was certain this thread was about real numbers, despite its name LINEAR ALGEBRA. Sorry for all the confusion VinnyCee!

 

[HallsofIvy]

Isn't |ax|=|a||x| by definition of a norm on a vector space?

That is part of the general definition. I suspect that this problem is based on
$$|ai+ bj+ ck|= \sqrt{a^2+ b^2+ c^2}$$
for R³ or
$$|ai+ bj|= \sqrt{a^2+ b^2}$$
for R². In that case radou's post is exactly what he needs.