Generally, I'm having quite troubles thinking about counter examples. So I would love if someone could guide me into finding the example for the following question by myself (and not just giving it away).
Let $$M=\{A=(a_{i,j})_{i,j\in N} | a_{i,j}\in \Bbb C, a_{i,j} \not=0 \text{ for only finitely many } i,j\}$$
[Please note that the matrices are infinite].
Define $\langle A,B\rangle =\text{tr}(AB^*)$.
I have managed to prove that this is indeed an inner product.
Now i'm trying to show that it is not a Hilbert space, by finding a cauchy's sequence that does not converge.
My "intuition" was to find a sequence of matrix such as in every matrix there would be only finite number of index' in which the value isn't 0, but it would have to "converge" to a matrix with infinte number of such index'.
I was trying to do it using matrix that would get values $1+1/n$, so it won't go to zero, and yet will somehow behave like a Cauchy sequence.
My two main idea (which were wrong) is $M_n =1+ 1/n$ on the diagonal until the $n$th place.
Any guidelines would be blessed. Thanks
Suppose that $A_n$ is the $\Bbb{N} \times \Bbb{N}$ matrix $A_n = \operatorname{diag}\{1, \frac{1}{2}, \frac{1}{4}, ..., \frac{1}{2^{n}}, 0, 0, ...\}$, then you can show that $\{A_n\}_{n=1}^{\infty}$ is a Cauchy sequence in $M$, but that it doesn't have a limit in $M$.
In fact, if you choose any sequence in $\ell_2(\Bbb{R})$ and build your matrices with the partial sequence along the diagonal in the same way as above, you'll be able to show the resulting sequence of matrices doesn't converge in $M$.