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

• VinnyCee
In summary: And "separate" rather than "separetely".In summary, the conversation discussed possible ways to prove the statement |a x| = |a| |x|, including using the Cauchy-Schwartz inequality and the definition of absolute value. It was suggested that treating different cases separately could lead to a proof, but it was not clear if this would be sufficient. The conversation also included some confusion and a clarification about the context of the problem.
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?

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

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

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.

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

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

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

If not, what would count as "proof"?

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

quasar987 said:
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?

Last edited:
quasar987 said:
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.

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

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?

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!

Galileo said:
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.

## 1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of vector spaces and linear transformations between them. It is used to solve systems of linear equations and to study geometric concepts such as lines, planes, and higher-dimensional analogues.

## 2. What is the significance of the equation |a x| = |a| |x|?

This equation is significant because it shows that the norm (magnitude) of a vector multiplied by a scalar is equal to the absolute value of the scalar multiplied by the norm of the vector. This is an important property in linear algebra and is often used in calculations and proofs.

## 3. Can you give an example of how this equation is used in real-world applications?

One example of how this equation is used in real-world applications is in computer graphics, where it is used to scale and rotate objects in 3D space. The equation ensures that the size of the object remains the same, even after it has been rotated or resized.

## 4. Is this equation valid for all real numbers a and x?

Yes, this equation is valid for all real numbers a and x. This is because the definition of the norm of a vector and the absolute value of a scalar are independent of the values of a and x.

## 5. How is this equation proven?

This equation can be proven using properties of vector operations and basic algebraic manipulations. It can also be proven geometrically by considering the lengths of vectors and the angle between them.

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