Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible.
I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^{n+1}(n-1)$. To prove this I do induction
n=2 we have the $A=\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ $\det(A)=-1$ and my formula gives me the same thing (-1)(2-1)=-1
Now assume if for $n \times n$ and $\det(A)=(-1)^{n+1}(n-1)$
Now to show for a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.
Best Answer
If you have studied eigenvalues and eigenvectors there is a very easy proof.
Let $A$ be your $n\times n$ matrix, with $n\ge2$. Then $A+I$ is the matrix consisting entirely of $1$s, which clearly has $n-1$ zero rows after row-reduction. Therefore $A$ has eigenvalue $-1$, repeated (at least) $n-1$ times, and since ${\rm trace}(A)=0$, the other eigenvalue is $n-1$.
Since every eigenvalue of $A$ is non-zero, the determinant of $A$ is non-zero, so $A$ is invertible.