# What is the Geometric meaning of vector norm in Rn n>3

My question is related to the length of the vector , Sorry it may seem stupid for you as i come from engineering background not mathematics background

For Vectors up to 3 dimensions (can be visualised) the following formula give the physical length from the origin to the tip of the vector (x1 , x2 , x3) (Pythagoras) length of vector in R3 but length of vectors above 3 dimensions is given by the general formula length of vector whose dimension above 3 , Now what is the Physical/Geometric interpretation of the second length formula ?

• If you assume that the Pythagorean theorem holds in $\mathbb{R}^n$ (that if $x,y$ are orthogonal then $||x+y||^2 = ||x||^2 + ||y||^2$) the formula follows by induction. Commented Mar 20, 2019 at 2:12
• It has the same meaning as in $\mathbb R^3$. To see this you can choose a a coordinate system with one of the axes lined up with your vector. Then its length is the same as it would be along the real number line. Commented Mar 20, 2019 at 2:14
• I am slightly confused. I'm not sure how to give a geometric interpretation of length. To me, it seems like saying the vector norm is length is the geometric interpretation already. Like, length is length regardless of how many dimensions there are. Is it possible for you to express what a geometric interpretation of length is in say two or three dimensions, so that we might try to give an $n$-dimensional analog?
– jgon
Commented Mar 20, 2019 at 2:14
• @JairTaylor could you please write an answer with the derivation (symbolic manipulation till reaching the second formula for distance) Commented Mar 20, 2019 at 2:22
• @jgon Don’t you measure lengths in three dimensions all the time in real life?
– amd
Commented Mar 20, 2019 at 4:32

## 2 Answers

The geometric meaning is the exact same; it is the length of the vector. For example, you could take a vector in $$\mathbb{R}^3$$ and map it, in a natural way, to a vector in $$\mathbb{R}^d$$. For example, $$(1,0,0) \in \mathbb{R}^3$$ could be viewed as $$(1,0,0,0,\ldots,0) \in \mathbb{R}^d$$.

• I understand your example , but the angle in R3 could be seen but in (1,0,0,0,...,0) where is the notion of angle Commented Mar 20, 2019 at 2:19
• The inner product (standard dot product) on $\mathbb{R}^d$ still allows us to define angles. Commented Mar 20, 2019 at 2:21

Length is a primitive notion - we have to have some definition before we can talk about it. But if you assume that $$|| \cdot ||$$ already exists in $$\mathbb{R}^n$$, and make the reasonable assumptions

1) that it obeys the Pythagorean theorem $$||x||^2 + ||y||^2 = ||x+y||^2$$ for $$x,y$$ orthogonal

2) that if $$e_i := (0, 0, \ldots, 0, 1, 0, \ldots, 0)$$ with a $$1$$ in the $$i$$-th place and $$0$$ elsewhere then $$||e_i|| = 1,$$

3) that $$||\lambda x|| = \lambda ||x||$$ for any scalar $$\lambda \geq 0$$ then you can derive the formula for it. These are all properties of $$|| \cdot ||$$ in $$\mathbb{R}^k$$ for $$k \leq 3$$, so it is natural to assume they should hold in $$\mathbb{R}^n$$.

You need not even define orthogonality in general - it's enough to assume that the $$e_i$$ are orthogonal. Note that (1) extends to any number of summands, e.g., $$||x+y+z||^2 = ||x + y||^2 + ||z||^2 = ||x||^2 + ||y||^2 + ||z||^2$$ if $$x,y,z$$ are orthogonal, or in general, $$||\sum_{i=1}^n v_i||^2 = \sum_{i=1}^n ||v_i||^2$$ when $$v_1, \ldots, v_n$$ are orthogonal - I'll let you see how to prove this with induction.

Note that for any $$(x_1, \ldots, x_n) \in \mathbb{R}^n$$, you get $$(x_1, \ldots, x_n) = x_1 \cdot (1, 0, \ldots, 0) + x_2 \cdot ( 0,1, \ldots, 0) + \ldots + x_n \cdot (0, \ldots, 0, 1)= \sum_{i=1}^n x_i e_i$$

and the summands $$x_i e_i$$ are orthogonal. So

$$||(x_1, \ldots, x_n)||^2 = ||\sum_{i=1}^n x_i e_i|| = \sum_{i=1}^n ||x_i e_i||^2 = \sum_{i=1}^n x_i^2 ||e_i||^2 = \sum_{i=1}^n x_i^2.$$

• in that link i.sstatic.net/aMuxV.png , if we follow the Pythagoras twice method then applying it third for example to a fourth vector orthogonal to the red vector (sum of first three) then that fourth vector in physical geometrical can be decomposed to the first three dimensions which means using this method of using Pythagoras several times will produce a dependant vectors for above 3 ... example (1,1,1) yields a length of sqrt(3) but for (1,1,1,1) what does sqrt(4)=2 represent geometrically ? Commented Mar 20, 2019 at 6:38
• simply I mean for (1,1,1,1) i go one step in x direction , then from there one further in y direction , from what i reached i go then one step in the z direction , then from what i reached a further step in any direction orthogonal to the vectorial sum of (1,1,1,1) (any direction in the plan orthogonal to the hypotenuse of (1,1,1)) now the problem i have is that the last fourth step can be decomposed in terms of the first three and how it then become independent of the first three ? Commented Mar 20, 2019 at 6:42