MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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
up vote 4 down vote accepted

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*}


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 $$ \sqrt{I + P(r,r,r-1)} = 1 + P(a,b,c). $$

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.