My question is just out of curiosity, I was trying to picture an orthonormal basis, in my mind picturing (using Cartesian coordinate system) three vectors, let's call them $x,y,z$ and as soon as I tried to add a fourth orthonormal vector, call it $t$, to the others to maintain a basis, the three dimensional static graph would not respect (graphically) the fact that all vectors are linearly independent (because the fourth was not orthonormal to all other vectors unless falling onto one of the first three vectors which would make it just a multiple of those vectors). So nothing astonishing, but is the only way to graph this only by drawing an $n^{th}$ dimension and just "thinking" it is orthonormal even if graphically it isn't? Is there a better system than the Cartesian one to represent this?
Thanks to anyone who can help solve this (if there is a solution)
Best Answer
A three-dimensional vector space, here $\mathbb{R}^3$ over the field $\mathbb{R}$ (i.e. 3D space), is such that by definition any basis is of size exactly 3. You cannot find any 4 linearly independent vectors.
In $n$-dimensional space, you have $n$ orthonormal vectors that form a basis. For instance, in $\mathbb{R}^n$ i.e. $n$-dimensional co-ordinates, such a basis is $e_1=(1,0,0,\dots,0),e_2=(0,1,0,\dots,0),e_3=(0,0,1,\dots,0),\dots,e_n=(0,0,0,\dots,1)$.
Graphically, if you're representing a 3D space on a 2D surface such as a piece of paper or a computer screen then you already just have to "think" that the vectors are orthonormal even though they are not literally so. Adding a fourth dimension has just the stumbling block that we lack the geometric intuition from the real world to picture 4D space, though representations like this 4D cube (tesseract) are not always so impenetrable.