The spectral radius of a $n\times n$ matrix

Question

I would like to know which is the spectral radius of this $n\times n$ matrix:

$$ \begin{matrix} 0 & 1 & . & . & . &1 \\ 1 & 0 & . & . & . &0 \\ . & . & . & & &. \\ . & . & & . & &. \\ . & . & & & . &. \\ 1 & 0 & . & . & . &0 \\ \end{matrix} $$

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I know that the spectral radius is the maximum eigenvalue, but I don't know how to calculate it in this matrix... I also know that if we've got a symmetric amtrix the spectral radius is $||A||_2$ but I neither know how to calculate this...

Answer 1

Your matrix has rank $2$, and in particular it can be written in the form $$ A = xy^T + yx^T, $$ where $x = (1,0,\dots,0)^T$ and $y = (0,1,\dots,1)^T$. Because $A$ has rank $2$, it has $0$ as an eigenvalue with algebraic multiplicity at least $n-2$; let $\lambda_1,\lambda_2$ denote the two possibly non-zero eigenvalues of $A$.

We can find the eigenvalues of $A$ by noting that the trace of a matrix is the sum of its eigenvalues. In particular, it is clear that $\operatorname{tr}(A) = 0$. Thus, we see that $$ \lambda_1 + \lambda_2 + 0 + \cdots + 0 = 0 \implies \lambda_1 = -\lambda_2. $$ On the other hand, we find that $$ A^2 = (xy^T + yx^T)^2 = xy^Txy^T + xy^Tyx^T + yx^Txy^T + yx^Tyx^T $$ Conclude that $$ \lambda_1^2 + \lambda_2^2 = \operatorname{tr}(A^2) \\= \operatorname{tr}[xy^Txy^T] + \operatorname{tr}[xy^Tyx^T] + \operatorname{tr}[yx^Txy^T] + \operatorname{tr}[yx^Tyx^T] \\= \operatorname{tr}[y^Txy^Tx] + \operatorname{tr}[x^Txy^Ty] + \operatorname{tr}[y^Tyx^Tx] + \operatorname{tr}[x^Tyx^Ty] \\= 2(x^Ty)^2 + 2(x^Tx)(y^Ty) = 2(n-1). $$ Conclude that the non-zero eigenvalues of $A$ are $\pm \sqrt{n-1}$, and the spectral radius is $\sqrt{n-1}$.

Answer 2

You can calculate $\Vert A\Vert_2$ explicitly.

Let $x=\pmatrix{x_1\\x_2\\\dots\\x_n}$ be a vector with norm equal to $1$. $Ax=\pmatrix{x_2+\dots+x_n\\x_1\\\dots\\x_1}$, and therefore:

\begin{eqnarray}\vert Ax\vert &=&\sqrt{(x_2+\dots+x_n)^2+(n-1)x_1^2}\\ &=&\sqrt{(x_2+\dots+x_n)^2+(n-1)(1-x_2^2-\dots-x_n^2)}\\ &=&\sqrt{(n-1)-\sum_{2\leq i<j\leq n}(x_i-x_j)^2}\\ &\leq&\sqrt{n-1} \end{eqnarray}

The equality holds for example if $x=\pmatrix{1\\0\\\dots\\0}$. Therefore $\Vert A\Vert_2=\sqrt{n-1}$.

Answer 3

We have $A=uv^T+vu^T=\pmatrix{u&v}\pmatrix{v&u}^T$ where $u=(1,0,\dots,0)^T$ and $v=(0,1,\dots,1)^T$. In general, $XY$ and $YX$ have the same multi-set of nonzero eigenvalues. Hence the nonzero eigenvalues of $A$ are those of $$ \pmatrix{v&u}^T\pmatrix{u&v}=\pmatrix{0&n-1\\ n-1&0}, $$ i.e. $\pm(n-1)$. The spectral radius of $A$ is therefore $n-1$.

Alternatively, note that $u$ and $w=v/\|v\|_2$ form a set of orthonormal vectors. Therefore $$ A=\|v\|_2(uw^T+wu^T) =\pmatrix{u&w}\left(\|v\|_2I_2\right)\pmatrix{w&u}^T $$ is an economic singular value decomposition. Hence $\|A\|_2=\|v\|_2=n-1$. However, as $A$ is symmetric, its spectral radius coincides with its spectral norm. Thus $\rho(A)=n-1$.