I have matrix $A \in M_{50}(\mathbb{R})$. This matrix has $A_{i,i} = 0$ for all $ 1 \le i \le 50$ and $A_{i,j} \in \{\pm1\}$ for all distinct $1 \le i, j \le 50$. I must show that $A$ is invertible.
In order to do so, I probably would want to show that $det(A)$ is non-zero. However, this is a 50 x 50 matrix, and surely, there must be another way to show that $det(A)$ is non-zero besides brute force computation. Someone told me to consider the image of $A$ in $M_{50}(F_{2})$, but I am not sure how to apply this into the proof. Can anybody clarify some things? Thank you.
Best Answer
Its determinant is an odd number, so is nonzero.
To see this, let $n$ be an even number (here $n$ is $50$) and let $M$ be a $n\times n$ matrix with zeros on the diagonal and $\pm 1$s elsewhere. Then $M\equiv J-I\pmod 2$ where $J$ is the all-one matrix. Therefore $\det A\equiv\det(J-I)=\det(I-J)\pmod 2$. The characteristic polynomial of $J$ is $$\det(tI-J)=t^{n-1}(t-nI)$$ since it has rank $1$ and the nonzero eigenvalue $n$. Therefore $\det(I-J)=1-n$, which is an odd number.