Hi,

This is probably a simple problem, but I just fell in love with that :D

Prove that for any $\displaystyle n\in\mathbb{N}$, the square matrix consisting in only 1's in the diagonal and n everywhere else is always invertible.

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- Jun 30th 2010, 09:25 AMMooInvertible matrix
Hi,

This is probably a simple problem, but I just fell in love with that :D

Prove that for any $\displaystyle n\in\mathbb{N}$, the square matrix consisting in only 1's in the diagonal and n everywhere else is always invertible. - Jun 30th 2010, 11:15 AMp0oint
Here is simple proof: The determinant of the matrix will always be $\displaystyle 1 \neq 0$ hence the matrix will be invertible.

- Jun 30th 2010, 11:16 AMBruno J.
It's not true for $\displaystyle n=1$... but for $\displaystyle n \neq 1$, here goes : it's trivial for $\displaystyle n=0$, and for $\displaystyle n>1$, the determinant $\displaystyle \mod n$ is equal to $\displaystyle 1$, so it can't be zero.

- Jun 30th 2010, 11:16 AMBruno J.
- Jun 30th 2010, 12:04 PMMoo
Yes, sorry, it's only for n>1.

And indeed, the simple way to prove it is to look at the matrix mod n :D

In *hard* times like these, I like algebra (Giggle) - Jun 30th 2010, 12:49 PMBruno J.
Algebra is great! :)

Here's a similar problem, perhaps a little harder.

If $\displaystyle n$ is an even positive integer, then the $\displaystyle n\times n$ matrix $\displaystyle A$ given by $\displaystyle A_{ij}=1 \ (i \neq j), \: \: A_{ii}=0$, is invertible. - Jun 30th 2010, 09:25 PMsimplependulum
The inverse is :

$\displaystyle -\frac{1}{n-1} I + \frac{n}{(n-1)(nm-n+1) } A $

where $\displaystyle [A]_{ij} = 1 ~~ 1 \leq i,j \leq m $ and $\displaystyle m $ is the size of the matrix .

To obtain the determinant , not only the residue ...

Consider the system of the linear equations :

$\displaystyle [ nA - (n-1)I ]x = y ~~ , ~ x = (x_1 ,x_2 ,...,x_m )^T ~,~ y = (0,0,0,...,0,1)^T $ and note that $\displaystyle nA - (n-1)I $ is your matrix , let it be $\displaystyle B_m$ ( If i read the problem correctly )

Let $\displaystyle S $ be the quantity $\displaystyle \sum x_i $ then we have

$\displaystyle nS - (n-1) x_i = 0 ~,~ 1 \leq i \leq m-1 ~~,~~ nS - (n-1)x_m = 1 $

$\displaystyle x_i = \frac{nS}{n-1} ~,~ 1 \leq i \leq m-1 ~~,~~ x_m = \frac{nS - 1}{n-1}$

$\displaystyle \frac{n(m-1)S}{n-1} + \frac{nS - 1}{n-1} = S $

$\displaystyle S = \frac{1}{nm - n + 1 } $

So $\displaystyle x_m = - \frac{n(m-1)-n+1}{(n-1)(nm-n+1) } = \frac{det(B_{m-1})}{det(B_m)} $

I believe it is true that $\displaystyle det(B_m) = (1-n)^{m-1}(nm-n+1)$ - Jul 1st 2010, 12:49 AMsimplependulum
- Jul 1st 2010, 01:51 AMtonio
- Jul 1st 2010, 02:49 AMsimplependulum
Oh , the matrix $\displaystyle A $ is based on what i defined in the previous post (#7) , sorry .