Invertible matrix

**Moo** · June 30th 2010, 09:25 AM

Hi,

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

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

**p0oint** · June 30th 2010, 11:15 AM

Here is simple proof: The determinant of the matrix will always be $1 \neq 0$ hence the matrix will be invertible.

**Bruno J.** · June 30th 2010, 11:16 AM

It's not true for $n=1$ ... but for $n \neq 1$ , here goes : it's trivial for $n=0$ , and for $n>1$ , the determinant $\mod n$ is equal to $1$ , so it can't be zero.

**Bruno J.** · June 30th 2010, 11:16 AM

Originally Posted by p0oint

Here is simple proof: The determinant of the matrix will always be $1 \neq 0$ hence the matrix will be invertible.

Beat me by a minute!

**Moo** · June 30th 2010, 12:04 PM

Yes, sorry, it's only for n>1.
And indeed, the simple way to prove it is to look at the matrix mod n

In *hard* times like these, I like algebra

**Bruno J.** · June 30th 2010, 12:49 PM

Algebra is great!

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

If $n$ is an even positive integer, then the $n\times n$ matrix $A$ given by $A_{ij}=1 \ (i \neq j), \: \: A_{ii}=0$ , is invertible.

**simplependulum** · June 30th 2010, 09:25 PM

The inverse is :

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

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

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

Consider the system of the linear equations :

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

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

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

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

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

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

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

I believe it is true that $det(B_m) = (1-n)^{m-1}(nm-n+1)$

**simplependulum** · July 1st 2010, 12:49 AM

Originally Posted by Bruno J.

Algebra is great!

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

If $n$ is an even positive integer, then the $n\times n$ matrix $A$ given by $A_{ij}=1 \ (i \neq j), \: \: A_{ii}=0$ , is invertible.