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

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

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...

User160
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    The columns are linearly dependent, so it's not hard to see the nullity = n - 2. The other 2 eigenvalues have to be opposite signs since the trace is zero. Just plug in a guess of the coefficients of your vector, and compute Av=alpha v, and you can derive that alpha = +/- something – mathmath8128 Jun 14 '20 at 20:33

3 Answers3

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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 $\sqrt{n-1}$.

Ben Grossmann
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  • One question, why is $tr(A^2)=2(x^{T}y)^2 +2(x^{T}x)(y^{T}y)$? – User160 Jun 14 '20 at 21:16
  • @User160 Sorry, I was a in a bit of a rush when I first posted this answer. See my latest edit for a justification. Note that I am using the cyclic property of trace. – Ben Grossmann Jun 15 '20 at 05:07
  • Since $x^Ty = 0$, $x^Tx = 1$, and $y^Ty = n-1$, don't you get $\text{tr}(A^2) = 2(n-1)$ instead of $2(n-1)^2$? Also, since $A$ is symmetric, the sum of the squares of the eigenvalues is $|A|_F^2 = 2(n-1)$, which is probably easier to compute. – JimmyK4542 Jun 15 '20 at 06:19
  • @JimmyK4542 Yes, that's correct. And that's a very good point about $|A|_F$ since that just amounts to counting the non-zero entries of $A$. – Ben Grossmann Jun 15 '20 at 06:21
  • Wouldn't be $tr[y^T xy^T x]+tr[x^T xy^T y]+tr[y^T yx^T x]+tr[x^T yx^T y]=2tr(x^T y)^2 +2tr[(x^Tx)(y^Ty)]$ ? – User160 Jun 15 '20 at 06:27
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    @User160 yes, but $(x^Ty)^2$ and $(x^Tx)(y^Ty)$ are both $1 \times 1$ matrices, so the "trace" is redundant. – Ben Grossmann Jun 15 '20 at 06:30
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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 $\Vert A\Vert_2=\sqrt{n-1}$.

anonymous67
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  • How do you know or why is it $x_1^2=(1-x_2^2-...-x_n^2)$? – User160 Jun 14 '20 at 21:17
  • The vector $x$ is supposed to have norm equal to $1$. In the definition of $\Vert A\Vert_2$, we have $\sup_{\vert x\vert=1} \vert Ax\vert$. And the norm of $x$ equal to 1 means $x_1^2+\dots+x_n^2=1^2=1$. – anonymous67 Jun 14 '20 at 21:33
  • Oh ok, and for last, how do you deduce that summatory? I mea, how do you know that what is inside the square is equal to (n-1) - that summatory? – User160 Jun 14 '20 at 21:39
  • You can prove it by induction. Here's how I think of it: the number of factor $x_k^2$ (for fixed $k$) is $n-2$, since each $x_k$ is paired $n-2$ times (each time with a $x_l,l\neq k$); the number of factor $x_kx_l$ is $2$; and these coefficients agree. In this case, I was lucky since we observe a clear square $\geq 0$. In general, you can use other method to find the minimal value (e.g. Lagrange's multipliers). – anonymous67 Jun 14 '20 at 21:50
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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 $\rho(A)=n-1$.

user1551
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