Orthonormal Basis In $l_1$ Norm

Question

If I am working on finite dimensional innerproduct space $\mathbb R^m$, and $\{v_{1},\dots,v_{m}\}$ be a orthonormal basis for $\mathbb R^m$, can I say that they are orthonormal in the sense of $l_1$ norm or $1$ norm of vector defined as $v_i=(v_{1i},v_{2i},\dots, v_{mi})^T$ as $\|v_i\|_1=\sum\limits_{i=1}^{m} |v_{ij}|$ and I define orthonormality as

$\langle v_i, v_j \rangle_{\text{1-norm}}=0\forall i\ne j$ and $\langle v_i, v_i \rangle_{\text{1-norm}}=1\forall i$?

Can I do this? Makes sense? Thanks for helping.

Answer 1

I think it works, but not for the mentioned example. But as he is defining the norm to be used is L1 instead of L2. so the unit distance will also be defined as per L1 norm and not L2. So a set of orthogonal vectors that have a unit length (unit as per L1 norm) will form the orthonormal basis (assuming it spans the space). For example in $R^2$ (1/2,1/2) and (-1/2,1/2) form the orthonormal basis as the L1 norm of each vector is 1. The L2 norm is not 1 but the author is defining it in terms of L1 and not L2.

Answer 2

As stated, this does not even work for $\mathbb{R}^2$ with the standard inner product. There $(\sqrt{2}/2,\, \sqrt{2}/2)$ has inner product with itself equal to one, but its 1-norm is not one. Let alone being equal to say this about a full orthonormal basis: it doesn’t even work for one element.