# Simple functions closed to Auerbach basis are linearly independent

$f_1,...,f_n \in L_1(\mu)$ is a Auerbach basis for an $n$-dimensional Banach space X,
i.e., $\{f_1,...,f_n\}$ is linearly independent, $\Vert f_i\Vert=1$, and there are $e_1,...,e_n$, linear functionals on $X$, such that $\Vert e_i\Vert=1$ and $e_i(f_j)=\delta_{ij}$
For each $i$, choose simple function $g_i$ such that $\Vert f_i-g_i\Vert_1<1/n^2$.
Can I conclude that $g_i$ are linearly independent?

## 1 Answer

This isn't really a Banach space problem as much as a linear algebra problem. The matrix formed by $e_i(f_j)$ is the identity matrix $I$. The matrix $e_i(f_j-g_j)$, call it $A$, has all entries less than $1/n^2$ in absolute value. And $e_i(g_j)$ is just $I-A$. We'd like this to be invertible.

To conclude that $I-A$ is invertible, it suffices to find a submultiplicative matrix norm $\|\cdot \|$ such that $\|A\|<1$, for then the Neumann series $\sum_{k=0}^\infty A^k$ converges and provides the inverse $(I-A)^{-1}$. Plenty of norms to choose from: max row sum, max column sum, Frobenius norm... they all have $\|A\|<1/n$.

• I don't understand why $I-A$ is invertible $\Rightarrow g_i$ are linearl independent. Can you explain more?
– CSH
Jul 26, 2017 at 23:03
• If the vectors are linearly dependent, then the columns of the matrix are linearly dependent.
– user357151
Jul 26, 2017 at 23:53
• Oh, I got it thx!!
– CSH
Jul 27, 2017 at 4:02