Something came to my mind as I was working on normed vector spaces over $\mathbb{R}$: are bases relevant when calculating certain norms?

For example, if we take the euclidean norm in $\mathbb{R} ^2$, we know that $\|(0,1)\|=1$. However one automatically assumes that we are talking about the canonical base here. But what if we take the basis $B=\{(1,0),(1,1)\}$? Then $(0,1)=(-1,1)_B$ but $\|(-1,1)_B\|=\sqrt2$, so we have two different "sizes" for essentially the same vector in $\mathbb{R} ^2$.

Now, I know that this example might make no sense because the euclidean norm is used with the canonical basis all the time, but I was just trying to make a point about whether the choice of a basis is always or sometimes relevant when calculating the norm of a vector, in any vector space. Surely not all norms depend on the coefficients of some linear combination, but when they do, how do we deal with bases?

I ask this mainly because of this thread : Proof that every finite dimensional normed vector space is complete, where the proof basically makes use of the equivalence between whatever norm is defined on V and the $\|\cdot\|_1$ norm. A random basis is taken, and it is noted that a Cauchy sequence is obtained for the coefficients of each vector in the sequence. These coefficients would vary when a different basis is taken, so how are we sure that there is nothing wrong or shady about this procedure?

I'm not sure if my question is clear, but I hope someone will get the idea!

  • $\begingroup$ You seem to be confused between vectors itself and their coordinates relative to a certain base. Coordinates fall together with vectors when viewed relative to the canonical base $\endgroup$ – user370967 Sep 13 '17 at 6:35

We do not assume that $\Vert (0,1) \Vert$ is calculated relative to canonical base. $(0,1)$ is just a vector. Your confusion is that you regard it as the coordinates of this vector relative to the canonical base.


You can define a euclidean norm for any basis: take a basis $b = \{b_1, b_2\}$ for $\mathbb{R}^2$ and define $\|\cdot\|_{2,b} : \mathbb{R^2} \to \mathbb{R}$ as:

$$\|x\|_{2,b} = \|\alpha b_1 + \beta b_2\|_{2,b} = \sqrt{\alpha^2 + \beta^2}$$

You can verify that $\|\cdot\|_{2,b}$ is a norm. Indeed, these norms are different for different bases (they are all eqiuvalent, though).

However, for any fixed basis $b$, its definition is basis independent since you can always write a formula for $\|\cdot\|_{2,b}$ using coordinates only:

Let $b = \{(x_1, y_1),(x_2, y_2)\}$ be a basis.

Then we have:

$$(x, y) = -\frac{x_2y+xy_2}{-x_2y_1+x_1y_2}\,(x_1,y_1)+\frac{x_1y-xy_1}{-x_2y_1+x_1y_2}\,(x_2,y_2)$$

So $$\|(x,y)\|_{2,b} = \sqrt{\left(\frac{x_2y+xy_2}{-x_2y_1+x_1y_2}\right)^2 + \left(\frac{x_1y-xy_1}{-x_2y_1+x_1y_2}\right)^2}$$

which is entirely basis independent.

  • $\begingroup$ So, equivalence of norms is basis independent as well? $\endgroup$ – Rlos Sep 13 '17 at 17:30
  • $\begingroup$ @CarlosGarcía Yes, all norms on a finite-dimensional space are equivalent. The constants may be different for each pair of norms, though. $\endgroup$ – mechanodroid Sep 13 '17 at 17:39

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