Is there an example of an injective infinite triangular matrix?

Question

Let $A$ be an infinite upper triangular matrix with complex entries and all diagonal entries nonzero, i.e.,

\begin{align} A=\left(\begin{matrix} a_{11}&a_{12}&a_{13}&\cdots\\ a_{21}&a_{22}&a_{23}&\cdots\\ a_{31}&a_{32}&a_{33}&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{matrix}\right) \end{align}

where $a_{ij}=0(j<i)$ and $a_{ii}\neq 0(\forall i)$. Moreover, assume that $A$ represents a bounded linear operator on an $\ell^p$-space, and for convenience, say, from $\ell^2(\mathbb N)$ to itself.

Although for the finitely dimensional case, such a triangular matrix must be bijective, the infinite-dimensional matrix $A$ does not have to be so. I previously asked a problem about it and a compact operator as an counterexample for it was formulated. What I would like to ask are:

• Is there any example that $A$ is injective and there are infinitely many subdiagonals of $A$ that are not eventually zero?

and moreover,

• Is there any condition (sufficient or necessary) on the entries of $A$ for $A$ to be injective?

Thanks in advance...

Answer 1

Although for the moment I don't have an answer to your first question, let's take a look at the second one:

Is there any condition (sufficient or necessary) on the entries of $A$ for $A$ to be injective?

I don't think there's much hope here (unless you require some additional properties of $A$, see below) and here's why: as also stated in this answer there exists an orthonormal basis of $\ell_2(\mathbb N)$ such that the left shift $S_L$ is upper triangular with diagonal $D(S_L)=\lbrace \frac12,\frac12,\ldots\rbrace$ but $S_L$ is not injective ($\operatorname{ker}(S_L)\neq \lbrace0\rbrace$ as is easy to see).

For a better answer the paper "Triangular Operators" (1990) by D. Herrero comes in handy - where I also took the left shift-example from. Herrero suggests that bitriangular operators might be of good use here. Note that an infinite complex matrix $A$ is called bitriangular if $A$ and its adjoint $A^\dagger$ are upper triangular (usually with respect to completely different orthonormal bases of $\ell_2(\mathbb N)$, unless $A$ is diagonal).

Theorem. Let $A\in\mathcal B(\ell_2(\mathbb N))$ (bounded linear operator) be bitriangular with non-zero diagonal so $0\notin D(A)$. Then $A$ is injective.

By Theorem 3.1 in said paper, for bitrinagular operators one has $D(A)=\sigma_p(A)$ where $\sigma_p(A)$ denotes the point spectrum (eigenvalues) of $A$. Now if $A$ is upper triangular with $0\notin D(A)$, then $0$ is not an eigenvalue of $A$ so $\operatorname{ker}(A)=\lbrace0\rbrace$ and thus $A$ is injective.

Answer 2

I guess I have come up with an example for my first question... Consider the matrix $A$ defined by

\begin{align} \left(\begin{matrix} A_1& 0_{1\times 2}&0_{1\times 3}&0_{1\times 4}&\cdots\\ 0_{2\times 1}&A_2&0_{2\times 3}&0_{2\times 4}&\cdots\\ 0_{3\times 1}&0_{3\times 2}&A_3&0_{3\times 4}&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{matrix}\right) \end{align}

where $A_n$ is an $n\times n$ block:

\begin{align} A_n=\frac{1}{n^2}\left(\begin{matrix} 1&1&1&\cdots&1\\ 0&1&1&\cdots&1\\ 0&0&1&\cdots&1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1 \end{matrix}\right)_{n\times n}. \end{align}

Then $A$ is an infinite matrix with all subdiagonals not eventually zero, and, since the sum of absolute square of its entries is $$ \sum_{i, j=1}^\infty|a_{ij}|^2\leqslant\sum_{n=1}^\infty\frac{1}{n^4}\cdot n^2<\infty, $$

$A$ is a compact operator (and of course bounded) on $\ell^2(\mathbb N)$.

Now that each diagonal block of $A$ is a finite upper triangular matrix with nonzero diagonal entries, if $Ax=0$ for some $x\in\ell^2(\mathbb N)$, then the corresponding block (of length $n$) in $x$ has to be a zero block. Therefore $x=0$, which implies that $A$ is injective.

Answer 3

Let $ T $ be the operator on $\mathcal l^2 (\mathbb N) $ given by the following:

$$T(e_0)=0~~~~~\text{and}~~~~ T(e_k)=e_{k-1}, ~~\text{if } k\geq 1 .$$ $T$ is called the bakward shift operator. Note that $T$can written in the following form $$\begin{pmatrix} 0 & 1 & 0 & 0 &\dots\\ 0 & 0 & 1 & 0 &\dots \\ 0 & 0 & 0& 1& \dots\\ \vdots&\vdots&\vdots&\ddots&\ddots \end{pmatrix} .$$

It can be shown that the spectrum of $T$ is $\bar{\mathbb D}.$ Now for any $\lambda\notin\bar{\mathbb D},$ consider the operator $$\lambda I+T .$$ It is a non-trivial uppertirangular infinite matrix on $\mathcal l^2 (\mathbb N) $ which is invertible and is a bounded linear operator.