Is infinite upper triangular matrix with nonzero entries on the diagonal invertible?

Question

Let's say we have an infinite square matrix where rows and columns are indexed by $\mathbb{N}$. We also know that every entry on the diagonal is nonzero and the matrix is upper triangular. Is it enough to conclude that our matrix is invertible?

It is true for finite matrices, but does it also hold for infinite matrices?

If not, what if we consider the case where only finite number of entries in each row and column are nonzero?

What I want to do is:

Let's say we have $$A(i, j) = \sum_{k \in \mathbb{N}} B(i, k) \cdot C(k, j)$$ for $i, j \in \mathbb{N}$ and we want to prove that there exists $D$ such that $$C(i, j) = \sum_{k \in \mathbb{N}} D(i, k) \cdot A(k, j).$$ Then we can rewrite it as a matrix product: $$A = B \cdot C.$$ Can we use the fact that $B$ is upper triangular and has nonzero diagonal to show that $$C = B^{-1} \cdot A$$ and $$C(i, j) = \sum_{k \in \mathbb{N}} B^{-1}(i, k) \cdot A(k, j)?$$

Answer 1

It depends on what you are expecting of infinite matrices. The following matrix $$\pmatrix{1&-1&0&0&\cdots\\0&1&-1&0&\\0&0&1&-1\\\vdots}$$ has an algebraic inverse $$\pmatrix{1&1&1&1&\cdots\\0&1&1&1&\\\vdots}$$ but it doesn't behave how you might expect. So for example, take the 'infinite' vector $$x=(1-\lambda)(1,\lambda,\lambda^2,\ldots)$$ then as $\lambda\to1$, $x\to0$ but $A^{-1}x\to(1,1,1,\ldots)$. If you're interested in this area, and want to make this rigorous, read on Hilbert spaces and Banach spaces.

Answer 2

First let us recall the following.

In the context of bounded linear operators, one says that $T$ is invertible if both $T$ and $T^{−1}$ are bounded linear operators.

Let

$$ D(T)=\lbrace \langle e_1,Te_1\rangle,\langle e_2,Te_2\rangle,\ldots \rbrace"="\lbrace T_{11},T_{22},\ldots\rbrace $$

be the diagonal entries of the upper triangular matrix $T$ with respect to some orthonormal basis $(e_n)_{n\in\mathbb N}$ of $\ell_2(\mathbb N)$. In finite dimensions we know that

$$ D(T)=\sigma_p(T)=\sigma(T) $$

no matter with regard to which orthonormal basis $T$ is upper triangular. Here $\sigma_p(T)$ is the point spectrum (eigenvalues) and $\sigma(T)$ is the whole spectrum of $T$. This is very useful since a non-zero diagonal directly tells us that $0$ is not in the spectrum so $T$ has to be invertible.

This is where the infinite dimensions come into play. Here we only have

$$D(T)\subset \sigma_p(T)\subset\sigma(T)$$

so the diagonal entries of upper triangular $T$ do not necessarily contain all the eigenvalues / elements of the spectrum so you can't directly read off if $T$ is invertible or not.

As an example, consider the left shift $S_L\in\mathcal B(\ell_2(\mathbb N))$ (bounded linear op. on $\ell_2$) which is upper triangular with respect to the standard basis where $D(T)=\lbrace 0,0,0,\ldots\rbrace$ as is readily verified. Note that $S_L$ has the open unit disk as point spectrum and the closed unit disk as whole spectrum so $S_L$ can't be invertible (in fact $S_L$ is surjective but not injective). However it is possible to find an orthonormal basis of $\ell_2(\mathbb N)$ such that $S_L$ is upper triangular and $0\notin D(S_L)=\lbrace \frac12,\frac12,\frac12,\ldots\rbrace$.

For the topic of triangular operators I recommend the paper "Triangular Operators" (1990) by D. Herrero - there I also took the example from.