Upper bound on the largest eigenvalue of a block matrix

Question

I would like to derive an upper bound on the largest eigenvalue of the following matrix in terms of the dimension of $X$ which is 5 and the dimension of the sub-matrix of zeros which is 2.
\begin{equation}
X=
\begin{pmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 \\
\end{pmatrix}
\end{equation}
In general, say we have an $n \times n$ matrix of ones. Define a matrix $X$ by subtracting an $m \times m$ matrix of ones with $m<n$ from the last $m$ columns and rows of the original matrix. Is there an upper bound on the largest eigenvalue of $X$ in terms of $m$ and $n$?

Answer 1

You can exactly find the nonzero eigenvalues in terms of $n$ and $m$. Note that there are two of them as this is a rank two matrix. We will use test vectors of the form $(ax,ax,\ldots,ax,x,\ldots,x)$, where there are $n-m$ copies of $ax$, $m$ copies of $x$, and where $x$ will be a nonzero indeterminate. The eigenvalue equations give \begin{gather} (n-m)ax+mx=\lambda ax\\ (n-m)ax=\lambda x. \end{gather} The second equation says we need $a=\lambda/(n-m)$. If we do this, the first equation solves to the following quadratic \begin{equation} \lambda +m=\lambda^2/(n-m)\iff \lambda^2-(n-m)\lambda-m(n-m). \end{equation} The quadratic formula yields \begin{equation} \lambda=\frac{(n-m)\pm\sqrt{(n-m)^2+4m(n-m)}}{2}. \end{equation} You can verify that this holds for your specific instance, yielding $\lambda_1=\frac{3+\sqrt{33}}{2}$.