An infinite hermitian matrix with each submatrix non singular

Question

Let $A$ be an hermitian infinite matrix with the property that $\forall n \in \mathbb{N}: A_{n \times n}$, the submatrix of order $n$ (rows and columns from $1$ to $n$), is non-singular.

My question: is it true that $A$ is non-singular ?

Thanks.

Answer 1

$A$ need not necessarily have an inverse. Let $$A=\begin{pmatrix}-1&1&1&1&\cdots\\1&1&0&\\1&0&1\\\vdots&&&\ddots\\1&&&&1&\\&&&&&\ddots\end{pmatrix}$$ Then every submatrix $A_{n\times n}$, with rows and columns from $1$ to $n$, is non-singular (with determinant $-n$).
If there were a matrix $B$ such that $BA=I$, then its first row would annihilate all the columns for $i>1$, so it would have to be of the form $a(-1,1,1,\ldots)$, yet multiplying out with the first column should give $1$, which is impossible.