# What can we do with a basis that we can't do without it?

All my life, I've learned to treat vectors as a set of 3 real numbers, which I multiply by a basis to get the actual vector: $$\langle a, b, c\rangle = ae_1+be_2+ce_3$$. As a programmer this is convenient because we can notate three real numbers in memory.

Recently, I've been looking at vectors and their kin from a different perspective. Like the classical way of viewing tensors, the vector is simply itself. There are operations I can do on it as a vector(like a dot product, or scalar product). These operations work perfectly without a basis. Its like everything I learned with a basis didn't need one at all!

What can I do with a vector if I have a basis that I can't do without one? The only operation I have found is "map a vector into 3 real numbers," which is useful to me as a programmer to encode a vector, but other than that, I'm having trouble finding what operations cannot be done without a basis.

• An example might be the projection to a basis-vector. – Strichcoder Mar 26 '19 at 3:28
• @Strichcoder Is that any more meaningful than a projection to any other sort of vector? – Cort Ammon Mar 26 '19 at 3:30
• If you have the projection onto all basis-vectors, then you also have a projection to any vector you wish. The thing is, that the projection depends on the basis you choose. – Strichcoder Mar 26 '19 at 3:36
• For your every-day use as a programmer, bases might not be that important. They are good for understanding the structures. Sometimes it is useful, however, to switch from one basis to another to have your objects in an easier form. Think of diagonilizing matrices. – amsmath Mar 26 '19 at 3:37
• @amsmath I'd think bases are extremely important to programmers! You would rarely try to store, say, a polynomial as a file structure in a computer, unless it was a coordinate vector. Similarly, you'd be more likely to deal with a linear map stored as a matrix than as a function/procedure. For example, the Gameboy Advance's rudimentary graphics hardware would allow for $2 \times 2$ matrices to be used in order to rotates/scale/skew sprites and certain background layers. – Theo Bendit Mar 26 '19 at 3:47

One example: with a fixed basis, you can find a canonical isomorphism from your vector space $$V$$ to its dual $$V^*$$ of linear functionals (I'm assuming finite dimensions here).

For a given basis $$(v_1, v_2, \ldots, v_n)$$ we can define a linear functional $$f_i$$ by defining its action on the basis, in particular, by defining $$f_i(v_i) = 1$$ and $$f_i(v_j) = 0$$ for $$j \neq i$$. In this way, we define the dual basis $$(f_1, \ldots, f_n)$$ for $$V^*$$ of $$(v_1, \ldots, v_n)$$.

This gives us a canonical isomorphism from $$V$$ to $$V^*$$, where we map $$a_1 v_1 + \ldots + a_n v_n \mapsto a_1 f_1 + \ldots + a_n f_n.$$ (Such a map is well-defined, since $$(v_1, \ldots, v_n)$$ is a basis.)

When you don't have a basis specified, there will obviously still be many such isomorphisms (e.g. pick any basis you want, and form an isomorphism as above), but there won't be an obvious, canonical isomorphism between the spaces.

• Any vector space with equal dimension will work as an example. – amsmath Mar 26 '19 at 3:40

You are implicitly using bases whenever you are performing some operation like dot product, cross product, scaling etc. on vectors.

The dot product is simply the sum of the products of coefficients for each base. You may not write the bases, but that doesn't mean you aren't using them. All operations deal with the coefficients of the bases.

EDIT: The application for bases is much more evident when you need to use alternate bases.

• The dot product also can be described as an area between the vectors. – Cort Ammon Mar 26 '19 at 3:30
• How about cross product, how do you know the resulting direction? The standard bases are such that you can get by without ever having to explicitly write them. And since your post talks about 3D vectors mainly, giving the direction is the most important task bases do. – Swapnil Rustagi Mar 26 '19 at 3:35
• By “direction” I’m guessing that you mean “orientation.” – amd Mar 26 '19 at 4:42
• @amd Yes, I meant orientation. – Swapnil Rustagi Mar 26 '19 at 6:32

Some operations like scalar product, cross product or general linear maps are independent of the choice of basis. There are some operations that are not independent. A simple example is given by the projection to a basis-vector.

If $$v=v_1\cdot e_1 +v_2 \cdot e_2$$ where $$v_1,v_2 \in \mathbb{R}$$, then the map $$\mathbb{R}^2 \to \mathbb{R}, v \mapsto v_1$$ depends on the basis $$\{e_1,e_2\}$$.

• Well, not necessarily. If you endow $\mathbb R^2$ with the standard Euclidean scalar product, this map can be completely specificed in terms of $v_1$ only. – amd Mar 26 '19 at 4:41