square root of a certain matrix [closed]

Hello, I'd like to know the square root of the following $n$ by $n$ matrix, for $n > 2$ and $r>0$:

$R_{ii}=r+1$ for $i < n$

$R_{ij}=r$ otherwise

The $2$ by $2$ case is given by

$\sqrt{R}=\frac{1}{d} \left[\begin{array}{cc} 1+r+\sqrt{r} & r \\\ r & r+\sqrt{r}\end{array}\right]$

where $d=\sqrt{1+2r+2\sqrt{r}}$.

Any thoughts? Many thanks.

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closed as too localized by Will Jagy, Nik Weaver, Federico Poloni, Chris Godsil, Denis SerreJun 18 '13 at 12:26

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Unless I'm mistaken, your matrix has minimal polynomial $(R-1)(R^2-(nr+1)R+r)=0$. From there it is easy to get its spectral form $E_1+\lambda_+E_++\lambda_-E_-$ where $\lambda_\pm=\frac12\left(nr+1\pm\sqrt{(nr+1)^2-4r}\right)$, and from there the square root $E_1+\sqrt{\lambda_+}E_++\sqrt{\lambda_-}E_-$. – Francois Ziegler Jun 18 '13 at 14:04
Here the eigenprojectors are, explicitly: $$E_1=\frac{R-\lambda_+}{1-\lambda_+}\frac{R-\lambda_-}{1-\lambda_-}$$ $$E_+=\frac{R-1}{\lambda_+-1}\frac{R-\lambda_-}{\lambda_+-\lambda_-}$$ $$E_-=\frac{R-1}{\lambda_--1}\frac{R-\lambda_+}{\lambda_--\lambda_+}$$ – Francois Ziegler Jun 18 '13 at 17:51

Let $P(a,b,c)$ be the $n\times n$ matrix where $$P(a,b,c)_{ij} = \begin{cases} a & \text{ if } i,j < n \\ b & \text{ if } i < n \text{ and } j = n \\ b & \text{ if } i = n \text{ and } j < n \\ c & \text{ if } i = j = n. \end{cases}$$

If I understand correctly, you want the square root of $I+P(r,r,0)$. You can check that $$(I+P(a,b,c))^2 = I + P((n-1)a^2+b^2+2a,(n-1)ab+bc+2b,(n-1)b^2+c^2+2c).$$ Thus, you just need to solve \begin{align*} (n-1)a^2+b^2+2a &= r \\\\ (n-1)ab+bc+2b &= r \\\\ (n-1)b^2+c^2+2c &= 0. \end{align*} Maple tells me that if we put \begin{align*} p &= \sqrt{1+(n-1)(r-r^2)} \\\\ q &= \sqrt{\frac{2p+2+(n-1)r}{(n-1)(n+3)}} \end{align*} then \begin{align*} a &= q + \frac{q-r-pq}{(n-1)r} \\\\ b &= q \\\\ c &= \frac{q-r-pq}{r}. \end{align*}

UPDATE:

The square root of $I+P(r,r,r-1)$ can be done similarly. If we put \begin{align*} s &= (1 - 2\sqrt{r} + nr)^{-1/2} \\\\ a &= rs + \frac{s - \sqrt{r}s - 1}{n-1} \\\\ b &= rs \\\\ c &= rs - \sqrt{r}s - 1 \end{align*} then $$\sqrt{I + P(r,r,r-1)} = 1 + P(a,b,c).$$

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I think the OP wants a square root of $I+P(r,r,r-1)$, not $I+P(r,r,0)$. – Francois Ziegler Jun 18 '13 at 14:11
Yes, I am interested in $I+P(r,r,r−1)$. Would this approach work for that case as well? (Unfortunately I am not a maple user). – user32851 Jun 18 '13 at 16:41
@unknown: I don't know. But what I sketched in my comment to your question above gives you the solution anyway. The point is that whenever the minimal polynomial factors without multiplicities, as in your case: $(R-1)(R-\lambda_+)(R-\lambda_-)=0$, then one has explicit formulas (which I have now added) for the eigenprojectors and hence a square root. – Francois Ziegler Jun 18 '13 at 17:56
Perfect, got it. Thanks so much! – user32851 Jun 18 '13 at 18:32