Vector Magnitude in dimensions > 4 I'm new to linear algebra, so please just dont blast me :(
I know the Pytagorean formula to calculate the magnitude of a vector in any vectorial space.
It's easy to grasp the meaning of the vector's magnitude in 2 and 3 dimensions, as the length of the displacement.
But, what's the general meaning of vector magnitude ? This in order to grasp the intuition behind vectors' magnitude in more than 3 dimension.
 A: Vector magnitude is still the length of the vector even in higher dimensions.
It might help to remember the Latin roots of three words here:


*

*vector is Latin for "carrier", like a ship or a horse, or "passenger"—something that goes for a ride. (Related to vehicle, a thing that you ride in.)

*magnitude is Anglicized Latin for "bigness". (Related to magnifier—a "biggifier".)

*direction comes from a Latin verb for "steer", i.e. to choose which way a vehicle travels.
When physicists wanted to name a mathematical quantity that would take a point for a "ride", simultaneously denoting both the distance and the direction, they borrowed the Latin words for these things. The magnitude of a vector is simply how far it carries a point.
It's easy to lose sight of this when the concepts have been abstracted to higher dimensions and even further levels of abstraction, but "being taken for a ride, a certain distance in a certain direction", is the underlying metaphor. 
A: There are lots of technical answers to your question, but if we take the concept of rotation in higher dimensions for granted, you can understand the length of a vector the same way you understand it in lower dimensions.
A vector is a one-dimensional object, you can always rotate it until it aligns with the x-axis, then its length is just what the usual length on the x-axis is. 
You can understand the formula $|\vec x|=\sqrt{\sum_i x_i^2}$, using multiple applications of Pythagorean theorem all in two-dimensional planes.
For example for a four-dimensional vector, there is a component of the vector along the fourth dimension. If you subtract that component, the remaining is a vector in the $x$-$y$-$z$-space. Let us call it $\vec u= \vec x - x_4\hat e_4$. You can rotate the space so that two vectors $\vec u$ and $\hat e_4$ are in your $x$-$y$-plane (or move the $u$-$e_4$-plane). Then by Pythagorean theorem, $|\vec x| = \sqrt{|\vec u|^2+|x_4\hat e_4|^2}= \sqrt{|\vec u|^2+x_4^2}$. This is just a right triangle in the two-dimensional space defined by $\vec u$ and $\hat e_4$, and $\vec u$ and $x_4\hat e_4$ are its sides. But you know that $|\vec u|^2= x_1^2+x_2^2+x_3^2$ (this is just the length in tree-dimension, you can deduce this by the same argument as above). Therefore, $|\vec x|= \sqrt{x_1^2+x_2^2+x_3^2+x_4^2}$.
A: The easiest way to intuit this (in finite dimensions) is to try to imagine your conception of "length of the displacement" in more than 3 dimensions.
Think about the way the concept of length in  2 dimensions extends to the concept of length in 3 dimensions when an extra orthogonal axis is added. In theory (though it's hard to visualize), we could add a 4th orthogonal axis and create a 4-dimensional space with distances and lengths.  Vector magnitude in 4 dimensions is length of the displacement in this new space.
A: In general terms the concept of length correspond to the norm which is a function that assigns a strictly positive length or size to each vector in a vector space; for the zero vector is assigned a length of zero.
On an n dimensional Euclidean space $\mathbb{R^n}$, the intuitive notion of length of the vector $x = (x_1, x_2, ..., x_n)$ is expressed by
$$\left\| \boldsymbol{X} \right\|_2 := \sqrt{x_1^2 + \cdots + x_n^2}$$
which gives the ordinary distance from the origin to the point X, as a consequence of the Pythagorean theorem.
A: If you have  4 variables $x,y,z,w$, and you move change these variables by $$ \delta x, \delta y, \delta z,\delta w $$ then$$ \sqrt {(   \delta x)^2+ ( \delta y)^2+ (\delta z)2+ (\delta w)^2}$$ is the magnitude of your change.  
A: For 1D-, 2D- and 3D- vectors we have some geometric intuition which tells us that 
$$|x|=\sqrt{x^2},\qquad\sqrt{x^2+y^2},\qquad \sqrt{x^2+y^2+z^2}$$
are the length of a $1$-, $2$-, or $3$-dimensional vector respectively. No such intuition exists for higher dimensions. But looking at these three initial examples should be enough to recognize a pattern. Why not just define
$$\sqrt{x^2+y^2+z^2+w^2}$$
as the length of a 4D-vector? And that's all. We just defined this for convenience. There is no reality we can compare with to see whether this definition is "correct" in any sense. It just turned out to be incredibly useful.
We call it length, but just for analogy. It is not really the "length" of anything in the real world. But you will appreciate the name because it allows you to think and reason about this magnitude as if it where something familiar. So for all purposes, in four and more dimensions, think of the length of a vector still as some kind of distance/displacement, even if you cannot picture it in your head. It can help to instead think of three dimensional analogons; they might be wrong, but more intuitive and the "wrongness" might cause not problems as long as you are aware of it.
