# Upper bound on the inverse of a Grammian matrix

I have been trying to find a reasonable upper bound on the following:

Given $n\in N$ and the Grammian matrix $A_n$ = (($f(i)$ , $f(j)$)) , $f(\lambda) = e^{\lambda t}$ for $0\le t \le 1$ and $1\le$$i,j\le$$n$ and $\lambda$ real, where ( $\cdot$ , $\cdot$ ) is the usual integral $L^2$ inner product of the vector space of continuous functions on the unit interval, then I need to find an upper bound on $$||A_n^{-1}||$$ The problem I am working on will be solved if I can show that the sequence of the above norms is bounded. Here the matrix is norm is $L_1$ norm i.e, the sum of absolute values of all entries.

My guess is that since the entries of $A_n$ grows exponentially, the inverse matrix should have sufficiently small entries; therefore, the sequence of norms is bounded. But I can not show it rigoriously.

Does anyone have a suggestion for me?

• $L^1$ isn't an inner product space. May 22, 2014 at 20:17
• Yeah, I meant to say $L^2$ norm. Thanks! May 22, 2014 at 20:34
• And can you please explain in more detail what continuous functions you mean when you write $ie^i$? You can edit your question to react to these comments. May 22, 2014 at 20:56

If you decompose your matrix as $$A_n=diag(e^1,...,e^n)\cdot \left(\!\!\left(\frac{1-e^{-(i+j)}}{i+j}\right)\!\!\right)_{i=1..n}^{j=1..n}\cdot diag(e^1,...,e^n)$$ then the inner matrix is close to the Hilbert matrix, which should help with finding estimates.