# Thread: Invertible matrix

1. ## Invertible matrix

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.

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

3. 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.

4. 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!

5. 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

6. 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.

7. 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 $det(B_m) = (1-n)^{m-1}(nm-n+1)$

8. 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.
Can't $n$ be an odd number $> 1$ ??

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

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

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

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

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

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

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

Tonio

10. Oh , the matrix $A$ is based on what i defined in the previous post (#7) , sorry .