On an operator norm inequality for two triangular matrices Let $n\ge3$. Create two copies $A$ and $B$ of the $n\times n$ lower triangular matrices of ones. Then set $A(2,1)=A(n,n-1)=B(n,n-1)=B(n,n-2)=0$ (with all other entries in the lower triangular parts still equal to $1$).
I want to show that the largest eigenvalue of $AA^T$ is greater than or equal to that of $BB^T$. Any hint to prove it?
 A: For large $n$, we can do this by perturbation theory. As my unperturbed matrix, I want to take $T=CC^*$, with $C$ being the lower triangular matrix with all entries equal to $1$. As in math110's answer here, we have that $T^{-1}=J$ is the Jacobi matrix with (almost) constant coefficients
$$
J= \begin{pmatrix}
2&-1\\
-1&2&-1\\
0&\ddots&\ddots&\ddots\\
\vdots&\cdots&-1&2&-1\\
0&\cdots&\cdots&-1&1
\end{pmatrix} .
$$
To find the eigenvalues, we must thus solve $-u_{k+1}-u_{k-1}+2u_k = E u_k$ (and the $2$ becomes a $1$ for $k=n$) with boundary conditions $u_0=u_{n+1}=0$. This we can do (almost) explicitly, by as usual building solutions from exponentials $u_k=\lambda^k$. We are interested in the smallest eigenvalue $E$. We find that $u_k=\sin\omega k$,
$$
E=\frac{\pi^2}{4n^2} + O(1/n^3) ,
$$
and $\lambda=e^{i\omega}$ solves $\lambda^2+1+(-2+E)\lambda=0$; in particular, $\omega=\pi/(2n)+O(1/n^2)$.
So $T$ has maximum eigenvalue $1/E=4n^2/\pi^2+O(n)$.
Now $BB^*=T-W$, and first order perturbation theory says that $1/E$ moves by $-\langle u, Wu\rangle/\|u\|^2$. Since $W$ is the symmetric matrix with the last three entries of its bottom row and final column equal to $1,2,2$, respectively, and everything else equal to zero, we have that
$$
\langle u, Wu\rangle = 8 + O(1/n) .
$$
Notice that $\|u\|^2\simeq n$, $\|W\|\simeq 1$, so this says that $BB^*$ has maximum eigenvalue $1/E-8/(cn) + O(1/n^2)$, and the corresponding normalized eigenvector satisfies $v=u/\|u\|+u_1+O(1/n^4)$, with $u_1=O(1/n^2)$ and $\langle u_1,u\rangle = 0$.
Similarly, $AA^*=T -V$, and $V$ has $1$'s in its second row and column and three more $1$'s in the bottom right corner. Thus
$$
\langle u, Vu \rangle = 3 + 2 + O(1/n) ,
$$
which is smaller than the corresponding term ($=8$) for $BB^*$, and this is good news.
We're not done yet though because $\simeq 1/n$ is smaller than what would have been possible based on $\|V\|\simeq n^{1/2}$, so the second order term could conceivably make a contribution $\simeq 1/n$ also.
To bypass this problem,  I instead want to check now that also $\langle v, AA^* v\rangle \ge \langle v, BB^* v\rangle$. The RHS I just computed approximately, up to errors $\lesssim 1/n^2$. Since $\langle u_1, Tu\rangle =0$, the LHS equals
$$
1/E - \langle u, Vu\rangle/\|u\|^2 - 2\langle u_1,Vu\rangle /\|u\| + O(1/n^2) ,
$$
and now the first two contributions, which we already worked out, do dominate everything else.
(We'd also have to check that lower lying eigenvalues can't overtake, but this is clear because the spacing is $\simeq n^2$.)
