Given that the eigenvalues of this matrix are all real, I need to find the spectral radius of this matrix
\begin{pmatrix} 2n-1&& n-1&& n\\ 1&& 2n-3&& 0\\ 1&& 0&& 1\end{pmatrix}
I used the Gersgorin Disc Theorem which gives that that the eigenvalues of the above matrix can lie in the intervals $(0,4n-2),(0,2), (2n-4,2n-2)$. However, I want to find a tighter upper bound of the spectral radius.
Are there any other theorems which can give a tighter upper bound of the above matrix?
Is there any other procedure to find an upper bound of the spectral radius of this matrix?
If someone could help, I would be grateful. Thank you.
Let $$ A = \begin{pmatrix} 2n-1&& n-1&& n\\ 1&& 2n-3&& 0\\ 1&& 0&& 1\end{pmatrix}. $$ Let $D$ be the diagonal matrix $$ D = \pmatrix{1\\&\sqrt{n}\\&&n}, $$ Consider the matrix $DAD^{-1}$, which is similar to $A$. We compute $$ DAD^{-1} = \pmatrix{ 2n-1&& \frac{n-1}{\sqrt{n}}&& 1\\ \sqrt{n}&& 2n-3&& 0\\ n&& 0&& 1 }. $$ The upper bound from the GCT applied to this matrix is $$ \max\{2n + \sqrt{n} - 1 - \frac 1{\sqrt n}, 2n + \sqrt{n} , n+1\} = 2n + \sqrt{n}. $$ More generally, you might find it useful to try this with different diagonal matrices $D$.
Another approach worth considering: we can write $A = Bn + C$, where $$ B = \pmatrix{2&1&1\\0&2&0\\0&0&0}, \quad C = \pmatrix{-1&-1&0\\1&-3&0\\1&0&1}. $$ From there, the spectral radius is bounded above by $$ r(A) \leq \|A\| = \|Bn + C\| \leq \|Bn\| + \|C\| = n \|B\| + \|C\|, $$ where $\|\cdot\|$ is any choice of submultiplicative matrix norm. For instance, taking $\|\cdot\|$ to be the spectral norm yields $\|B\| < 2.67$, $\|C\| < 3.25$, so that $$ \|A\| \leq 2.67n + 3.25. $$
On computing the best diagonal $D$:
Let $D$ be given by $$ D = \pmatrix{1\\&\alpha\\&&\beta}. $$ with $\alpha, \beta > 0$. We compute $$ D^{-1}AD = \begin{pmatrix} 2n-1&& \alpha(n-1)&& \beta n\\ \alpha^{-1}&& 2n-3&& 0\\ \beta^{-1}&& 0&& 1\end{pmatrix} $$ The resulting upper bounds on the spectral radius are the absolute row sums, namely $$ 2n - 1 + \alpha(n-1) + \beta n,\\ (2n - 3) + \alpha^{-1},\\ 1 + \beta^{-1}. $$ The best possible upper bound is the solution to the min-max problem $$ \min_{\alpha,\beta \geq 0} \max\{ 2n - 1 + \alpha(n-1) + \beta n, (2n - 3) + \alpha^{-1}, 1 + \beta^{-1} \}. $$