Jtdylktuy

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.

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 $ntimes n$ matrix with zeros on the diagonal and $pm 1$s
elsewhere. Then $Mequiv J-Ipmod 2$ where $J$ is the all-one matrix. Therefore
$det Aequivdet(J-I)=det(I-J)pmod 2$. 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.

Here is a proof along the lines of the hint you were given: considering $A in M_{50}(mathbb{Z})$, the image of $A$ in $M_{50}(mathbb{Z} / 2mathbb{Z})$ is the matrix with $1 + 2mathbb{Z}$ in each nondiagonal position and $0 + 2mathbb{Z}$ in each diagonal position. Let us call this image $B in M_{50}(mathbb{F}_2)$.

Now, we will show that $B$ is an invertible matrix. To see this, suppose we have $x in mathbb{F}_2^{50}$ with $Bx = 0$. Then from the first row, we have $x_2 + x_3 + cdots + x_{50} = 0$, so $x_1 = x_1 + x_2 + x_3 + cdots + x_{50}$. Similarly, the $i$th row gives $x_i = x_1 + x_2 + x_3 + cdots + x_{50}$ for each $i$. Therefore, $x_1 = x_2 = cdots = x_{50} = x_1 + x_2 + cdots + x_{50}$. But $x_1 = x_2 = cdots = x_{50} = 1 + 2mathbb{Z}$ will contradict the last equation; therefore, the only remaining possibility is that $x_1 = x_2 = cdots = x_{50} = 0 + 2mathbb{Z}$. We have thus shown that $B$ has trivial null space, so $B$ is invertible. (Note that for this last conclusion, we use the fact that $mathbb{F}_2$ is a field.)

This implies that $det(B) = 1 + 2mathbb{Z}$. Therefore, by the functoriality of the determinant, this implies that $det(A)$ is an odd integer; so $A$ is invertible as an element of $M_{50}(mathbb{R})$.