TITLE: Consider a matrix $A$ with $n=2m$ and $a_{ii}=a_{n+1-i,n+1-i} = a_{i,n+1-i}=a_{n+1-i,i} = i$ for $i=1,\dots, m$. Find the eigenvalues of $A$.
QUESTION [0 upvotes]: Consider a matrix $A$ with $n=2m$ and $a_{ii}=a_{n+1-i,n+1-i} =
a_{i,n+1-i}=a_{n+1-i,i} = i$ for $i=1,\dots, m$. Find $||A||_2$.

My attempt:
An example of such a matrix would be $(n=4, m=2)$, $$ \begin{pmatrix} 1&0&0&1\\0&2&2&0\\0&2&2&0\\1&0&0&1\end{pmatrix}.$$ The matrix is symmetric, therefore $||A||_2=\rho(A)$, where $\rho(A)=\max_i |\lambda_i|$ (where $\lambda_i$ are the eigenvalues of  $A$). The rank of this matrix is $m$, meaning that that the algebraic multiplicity of $0$ is at least $m$. Now, $\operatorname{tr}(A) = \sum_i \lambda_i = 2\sum_i^m i = m(m+1)$. 
The eigenvalues of the example that I gave are $0$ (algebraic multiplicity 2), $2$ (AM 1) and $4$ (AM 1). So, I would expect that in the general case $0, m$ and $m^2$ are eigenvalues, which $\rho(A)=m^2$ and therefore $||A||_2$. How can I show this rigorously? 
Thanks!

REPLY [1 votes]: If you reorder the basis by reversing the order of the last $m$ basis elements then the matrix is the tensor product of $\text{Diag}(1,2,\dots,m)$ and $\begin{bmatrix}1 &1\\1 &1\end{bmatrix}$. The latter matrix diagonalises to $\text{Diag}(2,0)$. The upshot is that your original  matrix is similar to $\text{Diag}(2,4,6,\dots, 2m,0,0,\dots, 0)$ and all is clear.
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