all norms on finite-dimensional vector spaces are equivalent


All norms on finite-dimensional vector spacesMathworldPlanetmath over R or C are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath (

A consequence of this is that there is only one norm induced topology on a finite dimensional vector space. This means that on such a vector space, we need not worry about what norm we use when we talk about convergence of a sequence of vectors in norm. So a standard use of this theoremMathworldPlanetmath is in continuity argumentsMathworldPlanetmath over finite dimensional vector spaces, and it allows you to pick the most convenient norm for your argument (the Euclidean normMathworldPlanetmath is not always very convenient).

This obviously is not true for infinite dimensional spaces, for example see the different Lp spaces ( Note that the reason all this works is because a unit sphere is compactPlanetmathPlanetmath in a finite dimensional vector space, while that is not true in an infinite dimensional one.


Any such finite-dimensional space is really just the same as n so we can talk about just those spaces. That is, any finite-dimensional vector space over or is isomorphic to n for some n (note that is just isomorphic to 2 as a vector space over ). To see this, just write any element of the space in of the basis and then define the isomorphismPlanetmathPlanetmathPlanetmath to take that basis to the standard basis in n and then extend linearly.

First let’s show that if two norms are equivalent on the unit sphere (all x such that x=1 with respect to some norm, for example the standard Euclidean norm) then they are equivalent everywhere. We can write any xn as a multiple of some scalar γ0 and a vector on the unit sphere, say x0, that is x=γx0. Then when suppose we have two equivalent norms, say a and b, on the unit sphere


So the norms are equivalent everywhere.

Suppose we are working with the 2-norm. Now we want to show that any other norm is a continuous functionMathworldPlanetmathPlanetmath with respect to the 2-norm. Take an arbitrary finite-dimensional space X and an arbitrary norm . Also suppose that {bi}1n is a basis of X and so an element xX may be written as x=1nxibi. Now given an ϵ>0, choose δ>0 such that x-y2<δ (the Euclidean distance is less then δ) implies that


In fact we can just choose δ to be the right side of the above inequalityMathworldPlanetmath. Now we note that the triangle inequalityMathworldMathworldPlanetmath immediately also yields the inequality |x-y|x-y. So


And so is a continuous function.

Suppose we are given two norms a and b, we know that they are both continuous functions with respect to the 2-norm. And so the function defined as


is a continuous function on the unit sphere (with respect to the 2-norm). This function is continuous except perhaps at 0, but we don’t care about the value at zero. On the unit sphere however f(x) is continuous and thus achieves a maximum and a minimum since the unit sphere is compact. Let’s call the minimum and maximum, α and β respectively. Then for any x on the unit sphere we have


And so the norms are equivalent on the unit sphere and thus as we shown above, everywhere. ∎

Title all norms on finite-dimensional vector spaces are equivalent
Canonical name AllNormsOnFinitedimensionalVectorSpacesAreEquivalent
Date of creation 2013-03-22 14:08:54
Last modified on 2013-03-22 14:08:54
Owner jirka (4157)
Last modified by jirka (4157)
Numerical id 15
Author jirka (4157)
Entry type Theorem
Classification msc 46B99