Injectiveness of an 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. Though any finite triangular matrix with nonzero diagonal entries must be invertible because of its nonvanishing determinant, it seems that for such an infinite matrix $A$, only its range can be easily seen. What I am wondering are:

• Is $A$ always injective (and if not is there any counterexample, and is there any condition under which $A$ is injective)?

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

Any help is greatly appreciated.

Answer 1

Here's the counterexample you were looking for, though this is only a partial answer to your question.

Let $x=(1/n)_{n\geq1}.$ If $(a_{i,j})_{j\geq 1}$ is defined so that $a_{i,i}=1/i,$ and $a_{i,j}=-(1/i^2j)\left(\sum_{j>i}1/j^2\right)^{-1},$ for all $j>i,$ then $Ax=0.$

Clearly each row of $A$ so defined belongs to $\ell^{2}(\mathbb{N}),$ with norm $$\|(a_{i,j})_{j\geq 1}\|^{2}=\frac{1}{i^{2}}+\left(\sum_{j>i}\frac{1}{j^{2}}\right)^{-2}\frac{1}{i^{4}}\sum_{j>i}\frac{1}{j^{2}}=\frac{1}{i^{2}}\left(1+\left(\sum_{j>i}\frac{i^2}{j^{2}}\right)^{-1}\right)\leq\frac{5}{i^{2}}.$$ Then for any $y\in\ell^{2}(\mathbb{N}),$ applying Cauchy-Schwarz to each term,$$\|Ay\|^{2}=\sum_{i\geq1}|\langle (a_{i,j})_{j\geq1},y\rangle|^{2}\leq\sum_{i\geq1}\|(a_{i,j})_{j\geq1}\|^{2}\|y\|^{2}\leq\|y\|^{2}\sum_{i\geq1}\frac{5}{i^{2}}=\frac{5\pi^{2}}{6}\|y\|^{2},$$ which proves $A$ is bounded.

Edit: The bound $\left(\sum_{j>i}\frac{i^{2}}{j^{2}}\right)^{-1}\leq 4$ is equivalent to $\sum_{j>i}\frac{i^{2}}{j^{2}}\geq\frac{1}{4},$ but this sum is always at least $\frac{i^{2}}{(i+1)^{2}},$ since all terms are nonnegative. This term is increasing with $i$, so it is minimized when $i=1,$ which gives $1/4$ as a lower bound.

Also note that $A$ is a compact operator, since if $A_n$ is the operator with only the first $n$ rows of $A$ and zero rows below, then $\|A−A_n\|^2\leq\sum_{i>n}\|(a_{i,j})_{j≥1}\|^{2}$ (proved similarly to the bound on $\|A\|$ above), and this sum goes to $0$ as $n\rightarrow\infty$, since $\sum_{i\geq1}\|(a_{i,j})_{j\geq1}\|^{2}<\infty$ as shown above. Since $A$ is the limit of finite-rank operators (in the operator norm topology), it is compact.