The Determinant of a Matrix with n+1 Ones: Is It Always -1, 0, or 1?

[mathmari]

Hey!

Let $A \in\mathbb{R}^{n\times n}$, $n\geq 3$ be a matrix with $n+1$ elements $1$ and the remaining elements are $0$. I want to show that $\det (A)\in \{-1, 0, 1\}$ and each of these $3$ possible values can occur.

Could you give me a hint how we could show that? I got stuck right now. (Wondering)

 

[I like Serena]

Hey!

Let $A \in\mathbb{R}^{n\times n}$, $n\geq 3$ be a matrix with $n+1$ elements $1$ and the remaining elements are $0$. I want to show that $\det (A)\in \{-1, 0, 1\}$ and each of these $3$ possible values can occur.

Could you give me a hint how we could show that? I got stuck right now. (Wondering)

Hey mathmari! (Smile)

What about induction?
We can include the 2x2 matrix for which it also holds true.

Did you know that we can evaluate any determinant like this:
$$\begin{vmatrix} a & b & c & d\\e & f & g & h\\i & j & k & l\\m & n & o & p \end{vmatrix}=a\,\begin{vmatrix} f & g & h\\j & k & l\\n & o & p \end{vmatrix}-b\,\begin{vmatrix} e & g & h\\i & k & l\\m & o & p \end{vmatrix}+c\,\begin{vmatrix} e & f & h\\i & j & l\\m & n & p \end{vmatrix}-d\,\begin{vmatrix} e & f & g\\i & j & k\\m & n & o \end{vmatrix}$$
(Wondering)

 

[mathmari]

Hey mathmari! (Smile)

What about induction?
We can include the 2x2 matrix for which it also holds true.

Did you know that we can evaluate any determinant like this:
$$\begin{vmatrix} a & b & c & d\\e & f & g & h\\i & j & k & l\\m & n & o & p \end{vmatrix}=a\,\begin{vmatrix} f & g & h\\j & k & l\\n & o & p \end{vmatrix}-b\,\begin{vmatrix} e & g & h\\i & k & l\\m & o & p \end{vmatrix}+c\,\begin{vmatrix} e & f & h\\i & j & l\\m & n & p \end{vmatrix}-d\,\begin{vmatrix} e & f & g\\i & j & k\\m & n & o \end{vmatrix}$$
(Wondering)

Base case:
For $n=2$ we have that $\begin{vmatrix} a & b \\ c & d\end{vmatrix}=ad-bc$.
Since $n+1=3$ elements are equal to $1$ we have that one element is $0$. That means that the result is either $1$ or $-1$.

For $n=3$ we have that $\begin{vmatrix} a & b & c\\ d & e & f \\ g & h & i\end{vmatrix}=a\,\begin{vmatrix} e & f \\ h & i\end{vmatrix}-b\,\begin{vmatrix} d & f \\g & i \end{vmatrix}+c\,\begin{vmatrix} d & e\\g & h \end{vmatrix}$.
Since $n+1=4$ elements are equal to $1$ we have that $9$ elements are equal to $0$.
If $a=b=c=0$ then the result is equal to $0$.
If at least one of the $a,b,c,d$ is non-zero, then we have from the previous case that the $2\times 2$ determinants will be equal to $1$ or $-1$. Therefore, the $3\times 3$ determinant will be equal to $-1$, $0$ or $1$.

Is this correct? Or could we improve the justification? (Wondering) Inductive hypothesis:
We suppose that $n\times n$ determinant will is equal to $-1$, $0$ or $1$. Inductive step:
The expansion of Laplace allows to reduce the computation of the $(n+1)\times (n+1)$ determinant to that of $n+1$ $(n\times n)$ determinants.
These $n+1$ determinants are equal to $-1$, $0$ or $1$.
We have $n+2$ elements that are equal to $1$ and the remaining ones are $0$.

How could we continue? I got stuck right now. (Wondering)

 

[I like Serena]

Let's distinguish 3 cases: the top row has only zeroes, it has exactly 1 one, or it has 2 or more ones... (Thinking)

 

[mathmari]

Let's distinguish 3 cases: the top row has only zeroes, it has exactly 1 one, or it has 2 or more ones... (Thinking)

If the top row has only zeroes, then the determinant of $A$ is equal to $0$.

If the top row has exactly 1 one, then the determinant is equal to determinant of the $n\times n$ submatrix, of which the coefficient is $1$. From the inductive hypothesis we get that this will be equal to $-1$, $0$ or $1$, right? (Wondering)

If the top row has 2 or more ones, then the determinant of $A$ is equal to the determinant of the 2 or more $n$-submatrices. These, from the inductive hypothesis, will be equal to $-1$, $0$ or $1$. How do we know that the result is not $2$ for example? (Wondering)

 

[I like Serena]

Isn't the inductive hypothesis that an nxn matrix with n+1 ones is -1,0, or 1?
What if an nxn matrix has n ones? Or less than n ones?