# Invertible matrix

• Jun 30th 2010, 09:25 AM
Moo
Invertible 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 AM
p0oint
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 AM
Bruno 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 AM
Bruno J.
Quote:

Originally Posted by p0oint
Here is simple proof: The determinant of the matrix will always be $\displaystyle 1 \neq 0$ hence the matrix will be invertible.

Beat me by a minute! :)
• Jun 30th 2010, 12:04 PM
Moo
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 PM
Bruno 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 PM
simplependulum
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 AM
simplependulum
Quote:

Originally Posted by Bruno 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.

Can't $\displaystyle n$ be an odd number $\displaystyle > 1$ ??

The inverse is $\displaystyle \frac{1}{n-1} A - I$
• Jul 1st 2010, 01:51 AM
tonio
Quote:

Originally Posted by simplependulum
Can't $\displaystyle n$ be an odd number $\displaystyle > 1$ ??

The inverse is $\displaystyle \frac{1}{n-1} A - I$

The above doesn't work even for $\displaystyle n=2$: $\displaystyle A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\Longrightar row A^{-1}=A\neq A-I$ .

Spoiler:
Let's try to find out what conditions must be met so that $\displaystyle A\underline{x}=\underline{0}\,,\,\,\underline{x}=\ begin{pmatrix}x_1\\x_2\\\ldots\\x_n\end{pmatrix}$ :

$\displaystyle \begin{pmatrix}0&1&\ldots&1&1\\1&0&1&\ldots &1\\\ldots&\ldots&\ldots&\ldots&\ldots\\1&1&\ldots &1&0\end{pmatrix}$ $\displaystyle \begin{pmatrix}x_1\\x_2\\\ldots\\x_n\end{pmatrix}= \begin{pmatrix}x_2+x_3+\ldots+x_n\\x_1+x_3+\ldots+ x_n\\\ldots\\x_1+x_2+\ldots +x_{n-1}\end{pmatrix}$.

If we equal the above to the zero vector and solve we'll find at once that it must be that $\displaystyle x_1=x_2=\ldots =x_n=0$ , which means $\displaystyle A$ is 1-1 as linear transformation and thus regular.

Tonio
• Jul 1st 2010, 02:49 AM
simplependulum
Oh , the matrix $\displaystyle A$ is based on what i defined in the previous post (#7) , sorry .