# equivalence of norms arising from inner products

It's well known that all norms on a finite dimensional vector space are equivalent, but does restricting to norms arising from inner products make a meaningful difference to how someone could go about proving this result? In other words, can the fact that the norms come from inner products be exploited for an alternative argument?

My motivation for this question comes from Axler's 'Linear Algebra done right' (3rd Ed.). Specifically, question 6.B.12 asks for a proof that two inner products on a finite dimensional vector space produce equivalent norms. Does he intend students to essentially prove the full result, or is there an easier proof for the special case?

• If you are familiar with matrix representations of bilinear forms, diagonalization of bilinear forms, and spectral theory, then there is a more clever proof. (Simply apply the spectral theorem to the matrix representation of the inner product, which is necessarily Hermitian positive definite.) Commented Dec 3, 2017 at 10:45
• That's a neat proof, but I don't think he intends that as that material has not been covered by that point in the book. Commented Dec 3, 2017 at 10:49
• Hence why I prefaced it with the "if". Using all that is like the mathematical equivalent of killing a fly with a missile. It's total overkill, but it may still be of some interest to you which is why I left it in a comment rather than an answer. I'm not familiar with a more elementary proof (that is specific to inner products) Commented Dec 3, 2017 at 10:51
• I appreciated your comment. Thank you. Commented Dec 3, 2017 at 10:52
• My intention with this exercise is that only topics that have already been introduced in the book should be used in the solution. Thus the spectral theorem is not intended to be used. Also not intended to be used is the harder result that all norms (not just norms coming from an inner product) are equivalent on a finite-dimensional vector space. Tools available at this stage of the book include: representation of a vector as a linear combination of orthonormal basis vectors, Triangle Inequality, Cauchy-Schwarz Inequality. Commented Dec 4, 2017 at 1:30

The answer is yes. There is a significantly simpler way to proving the restricted result. The idea is to choose a basis for the vector space that is orthonormal wrt one of the inner products. Given a vector $v$ expressed in terms of the chosen basis, the norm of $v$ wrt the corresponding inner product can just be written down. The result then follows from the triangle inequality, and the fact that if $n$ is a positive integer, and if $x_1,\ldots,x_n\in\mathbb{R}$, then $(x_1+\ldots + x_n)^2\leq n(x_1^2+\ldots + x_n^2)$, which is proven using Cauchy-Schwarz as exercise 6.A.12 in the same book.