TITLE: Trace norm of a triangular matrix with only ones above the diagonal
QUESTION [8 upvotes]: For $n\in\mathbb N^*$, we consider the triangular matrix
$$
T_n = 
\begin{pmatrix}
1 & \cdots & 1 \\
  & \ddots & \vdots \\
0 &  & 1
\end{pmatrix}
\in M_{n,n}(\mathbb R)
\,.
$$
The trace norm of $T_n$, that is the sum of the singular values of $T_n$, is denoted by $\|T_n\|_{\text{Tr}}$.
Is it true that
$$
\sup_{n\in\mathbb N^*} \Big\{\frac{1}{n}\|T_n\|_{\text{Tr}}\Big\} < \infty \,?
$$

EDIT
Is it true that
$$
\sup_{n\in\mathbb N^*} \Big\{\frac{1}{n\log(n)}\|T_n\|_{\text{Tr}}\Big\} < \infty \,?
$$
An equivalent definition of the trace norm is $\|T_n\|_{\text{Tr}}:=\text{Tr}[\sqrt{T_n^T T_n}]$, where the square root $\sqrt{A}$ of a nonnegative matrix $A$ is the only nonnegative matrix such that $\sqrt{A}^2=A$. (And by $A$ nonnegative I mean $⟨u,Au⟩\geq0$ for all vector $u\in\mathbb R^n$).

EDIT 2
One can explicitly compute the singular values of $T_n$.
$$T_n^{-1} =
\begin{pmatrix}
1 & -1 & & 0 \\
  & \ddots & \ddots & \\
 & & \ddots & -1 \\
0 & & & 1
\end{pmatrix}
\in M_{n,n}(\mathbb R)
\,.
$$
The singular values $\sigma_1,\dots,\sigma_n$ of $T_n$ are related to those, $\lambda_1,\dots,\lambda_n$, of $T_n^{-1}$ through $\sigma_j=\lambda_j^{-1}$.
It is easier to compute the singular values of $T_n^{-1}$ because the eigenvalues $\mu_j=\lambda_j^2$ of
$$A_n = (T_n^{-1})^* T_n^{-1} =
\begin{pmatrix}
  1    &    -1  &        &        &     0  \\
 -1    &    2   &  -1    &        &        \\
       &    -1  & \ddots & \ddots &        \\
       &        & \ddots & \ddots &    -1  \\
  0    &        &        &    -1  &     2
\end{pmatrix}
\,,$$
can be computed explicitly (as for the discrete laplacian).
Eigenvalues of $A_n$
First, $A_n$ is real symmetric, hence it can be diagonalized in an orthonormal basis, and its eigenvalues $\mu_j$ are real. Then using, say, Gershgorin's circle theorem, the eigenvalues are in the interval $[0,4]$.
If $\psi=(\psi_1,\dots,\psi_n)^T$ is an eigenvector of $A_n$ associated with the eigenvalue $\mu$, then
\begin{align}
\psi_2 & = (1-\mu)\psi_1 & (1) \\
\psi_3 & = (2-\mu)\psi_2 - \psi_1 & (2)\\
 & \vdots \\
\psi_{j+2} &=  (2-\mu)\psi_{j+1} - \psi_j & (j)\\
 & \vdots \\
\psi_n & = (2-\mu)\psi_{n-1} - \psi_{n-2} & (n-1)\\
(2-\mu)\psi_n &= \psi_{n-1} & (n)
\end{align}
From Eq. $(2)$ to $(n-1)$, one can see that $\psi_j$ is linear, recursive sequence of order two. Since the roots of the polynomial $X^2+(\mu-2)X+1$ are
$$\frac{2-\mu\pm i \sqrt{\mu(4-\mu)}}{2}=e^{\pm i\theta}$$
with $\theta\in [0,\pi]$ and $\cos(\theta)=1-\frac{\mu}{2}$, $\psi_j=\Re(a e^{i(j-1)\theta})$ with $a$ a complex number. Up to a (real) normalization factor $\psi_j=\Re(e^{i(\varphi+(j-1)\theta)})$ for some $\varphi\in\mathbb R$.
Using $\mu = 2-e^{i\theta}-e^{-i\theta}$ and Eq.(1) and (n), yields
\begin{align}
e^{i(\varphi+\theta)}+e^{-i(\varphi+\theta)}&=(e^{i\theta}+e^{-i\theta}-1)(e^{i\varphi}+e^{-i\varphi}) \\
(e^{i\theta}+e^{-i\theta})(e^{i(\varphi+(n-1)\theta)}+e^{-i(\varphi+(n-1)\theta)})&=e^{i(\varphi+(n-2)\theta)}+e^{-i(\varphi+(n-2)\theta)}
\end{align}
i.e.
\begin{align}
\cos(\varphi-\theta)&=\cos(\varphi) & (1)'\\
\cos(\varphi+n\theta)&=0 & (n)'
\end{align}
From $(1)'$, either $\theta=0$ or $\varphi = \frac{\theta}{2}+k\pi$. $\theta=0$ would give $\mu=0$ which is excluded since $A_n$ is an invertible matrix. Hence using $\varphi = \frac{\theta}{2}+k\pi$ and $(n)'$:
$$(n+\frac{1}{2})\theta=(k+\frac{1}{2})\pi$$
Using that $\theta\in[0,\pi]$, we get that $\theta\in \Big\{\frac{j-\frac{1}{2}}{n+\frac{1}{2}} \pi \mid j=1,\dots,n\Big\}$. And in fact each of these values yields an eigenvalue and an eigenvector. The corresponding eigenvalues are $\mu_j=4\sin^2\Big(\frac{j-\frac{1}{2}}{n+\frac{1}{2}} \frac{\pi}{2}\Big)$, $ j=1,\dots,n$. 
Trace Norm of $T_n$
The singular values of $T_n$ can now be deduced:
$$\sigma_j=\frac{1}{2\sin\Big(\frac{j-\frac{1}{2}}{n+\frac{1}{2}} \frac{\pi}{2}\Big)}\,,\quad  j=1,\dots,n$$
and the trace norm is
$$ \|T_n\|_{\text{Tr}}=\frac{1}{2}\sum_{j=1}^n \frac{1}{\sin\Big(\frac{j-\frac{1}{2}}{n+\frac{1}{2}} \frac{\pi}{2}\Big)} \,.$$
Using that $\sin x\geq \frac{2}{\pi}x$ on $[0,\frac{\pi}{2}]$ one gets the upper bound:
$$ \|T_n\|_{\text{Tr}}\leq \frac{n+\frac{1}{2}}{2}\sum_{j=1}^n \frac{1}{j-1/2}\leq \frac{n+\frac{1}{2}}{2} \Big(\frac{1}{2}+\ln(2n+1)\Big)\,,$$
which implies that
$$
\limsup_{n\in\mathbb N^*} \Big\{\frac{1}{n\log(n)}\|T_n\|_{\text{Tr}}\Big\} \leq \frac{1}{2} \,.
$$
Actually one also has a lower bound
$$ \frac{n+1/2}{\pi}\Big(\ln(\tan(\frac{\pi}{4}))-\ln(\tan(\frac{\pi}{4(n+1/2)}))\Big)
\leq \frac{n+1/2}{\pi} \int_{\frac{\pi}{2n+1}}^{\frac{\pi}{2}} \frac{dx}{\sin(x)} 
\leq \|T_n\|_{\text{Tr}} \,,$$
which implies that
$$
\frac{1}{\pi} \leq \liminf_{n\in\mathbb N^*} \Big\{\frac{1}{n\log(n)}\|T_n\|_{\text{Tr}}\Big\} \,.
$$

REPLY [6 votes]: In Loewner ordering, we have
$$
T_n^{-1}(T_n^{-1})^\top
=\pmatrix{2&-1\\ -1&\ddots&\ddots\\ &\ddots&2&-1\\ &&-1&1}
\preceq\pmatrix{2&-1\\ -1&\ddots&\ddots\\ &\ddots&2&-1\\ &&-1&2}=P.
$$
Using the spectral formula for tridiagonal Toeplitz matrices, the eigenvalues of $P$ are given by $2+2\cos\left(\frac{k\pi}{n+1}\right)=4\cos^2\left(\frac{k\pi}{2(n+1)}\right)$ with $k=1,2,\ldots,n$. Therefore, the $k$-th smallest singular value of $T_n$ is bounded below by $\frac12\sec\left(\frac{k\pi}{2(n+1)}\right)$ and
\begin{align}
\frac1n\|T_n\|_{\text{Tr}}
&\ge\frac1n\sum_{k=1}^n\frac12\sec\left(\frac{k\pi}{2(n+1)}\right)\\
&\ge\frac1\pi\int_0^{n\pi/2(n+1)}\sec x\,dx\\
&=\frac1\pi\ln\left(\sec\frac{n\pi}{2(n+1)} + \tan \frac{n\pi}{2(n+1)}\right),
\end{align}
which is unbounded when $n\to\infty$.