Correct my simple error (property of scalar multiplication)

[iScience]

say i have some vector $\vec{v}$ multiplied by a scalar k.

the norm of $\vec{v}$ would be just $||\vec{v}||$ and the norm of $k\vec{v}$ is claimed to be $|k|||\vec{v}||$ ie, $||k\vec{v}||=|k|||\vec{v}||$ ie the sign of the constant is irrelevant.

when i work it out..

$$||k\vec{v}||=\sqrt{(kv_{1})^2+(kv_{2})^2+(kv_{3})^2+...}$$
$$=\sqrt{k^2(v_{1})^2+k^2(v_{2})^2+k^2(v_{3})^2+...}=\sqrt{(k)^2[(v_{1})^2+(v_{2})^2+(v_{3})^2+...]}$$

from this point, yes i could say that since all k values are squared, the square root must be a positive value. but take k to be a negative number (-a) and say we don't actually compute the square of k and we just leave it in $(-a)^2$ form. we can do the following

(continued from previous)

=$$\sqrt{(-a)^2}\sqrt{(v_{1})^2+(v_{2})^2+(v_{3})^2+...}$$

in which case, the $\sqrt{}$ and the $(-a)^2$ can cancel out which would then return our original -k value which is negative. (where -k=-a)

i'm pretty sure there's an error some where in my line of reasoning though, could someone please find it?

 

[Mark44]

I don't know if you're aware of it, but there were four copies of this post where it looked like you were trying different things in LaTeX. I deleted all but this one, which was the best of the lot.

say i have some vector $\vec{v}$ multiplied by a scalar k.

the norm of $\vec{v}$ would be just $||\vec{v}||$ and the norm of $k\vec{v}$ is claimed to be $|k|||\vec{v}||$ ie, $||k\vec{v}||=|k|||\vec{v}||$ ie the sign of the constant is irrelevant.

when i work it out..

$$||k\vec{v}||=\sqrt{(kv_{1})^2+(kv_{2})^2+(kv_{3})^2+...}$$
$$=\sqrt{k^2(v_{1})^2+k^2(v_{2})^2+k^2(v_{3})^2+...}=\sqrt{(k)^2[(v_{1})^2+(v_{2})^2+(v_{3})^2+...]}$$

(from this point, yes i could say that since all k values are squared, the square root must be a positive value. but take k to be a negative number and say we don't actually compute the square of k and we just leave it in $$(-k)^2$$ form. we can do the following)

=$$\sqrt{k^2}\sqrt{(v_{1})^2+(v_{2})^2+(v_{3})^2+...}$$

in which case, the $\sqrt{}$ and the $k^2$ can cancel out which would then return our original -k value which is negative.

No.
These operations don't just cancel out. $\sqrt{x^2} = |x|$.

Think about it. Although $\sqrt{4^2} = 4$, it's NOT true that $\sqrt{(-4)^2} = -4$.

i'm pretty sure there's an error some where in my line of reasoning though, could someone please find it?

 

[ehild]

The norm of a vector is defined to be non-negative. So the norm of the vector -3 i is 3 and not -3.

The real operation √x is defined for non-negative x values. √x is the non-negative quantity which has the square equal to x:(√x)²=x. √x ≥0.

If a = -2, for example, and x=(-2)², √x=√4=2.

√(x²)=|x| (the square root does not have memory )

ehild

 

[iScience]

well, if this is expressed in the following way:

$$\sqrt{x^2}=x^\frac{2}{2}$$

are you guys saying that if i have any fractional exponent $$\frac{x}{y}$$

where $$x\in even numbers , y\in\Re$$

that i can't simply simplify the exponent if y is an even number? if x is an even number then the quantity must be positive?

because i have done homework problems in the past where i have a quantity such as this

$$(-x)^\frac{2}{6}$$ and i reduced this to $$(-x)^\frac{1}{3}$$, solved the problem and have gotten them correct. how is this any different from my original inquiry??

 

[ehild]

well, if this is expressed in the following way:

$$\sqrt{x^2}=x^\frac{2}{2}$$

You can not express it that way.

ehild