Determinant of the anti-diagonal square matrix filled with 1's.

Question

So as the title suggests, we have an nxn matrix with the coefficients on the anti-diagonal as 1 and all others 0.

I've been trying to use the Laplace expansion on the 1st column:

$$det(A) = \sum_{i=1}^{n}(-1)^{i+1}\cdot a_{i1} \cdot det(A_{i1})$$

So the first term looks like:

$$det(A) =(-1)^{n+1} \cdot{det(A_{n1})}$$

Because only the last term in the expansion is nonzero.

Then using the expansion on this det:

$$det(A_{n1})= (-1)^{n}\cdot det(A_{(n-1)1}) $$

So:

$$det(A) = (-1)^{2n+1}\cdot det(A_{(n-1)1})$$

But then, wouldn't $A_{(n-1)1}$ be a matrix with all zeros in the first columns, and so the determinant of this matrix would be zero, and then $det(A) = 0 $

I don't feel like this is correct.

If anyone can show me the correct way to use the expansion for this matrix, that would be appreciated.

Answer 1

Apply the rule that says that swapping two columns (or two rows) changes the sign.
Then it is easy to deduct that the determinant for a $n \times n$ "exchange matrix" is $ \left( { - 1} \right)^{ \,\left\lceil {{{n - 1} \over 2}} \right\rceil } $

So $n=1$ no swap, $n=2$ one swap, $n=3$ one swap ( 1st with 3rd returns I), $n=4$ two swaps (1-4,2-3), etc.
If $n$ is even you exchange each couple $(1,n), (2,n-1),\cdots$, if $n$ is odd the central column remains fixed: no of exchanges $\left\lfloor{n/2}\right\rfloor = \left\lceil {(n-1)/2 }\right\rceil$.

Answer 2

The only nonzero product is $1$ by going down the antidiagonal, so det is not $0.$ all one needs is the sign, and that comes from the permutation associated to the antidiagonal. That may be $\pm 1$ depending on the matrix size. Try a few small $n$ as in $n \times n$ and look for the pattern.

Edit: As noted by DanielV one can apply the Leibniz expansion. If $A_n$ is the $n \times n$ antidiagonal matrix, then the sign associated to the $1$ of row $1$ column $n$ is $(-1)^{n+1}=(-1)^{n-1}.$ Then since $\det(A_1)=1=(-1)^0,$ one has $\det(A_n)$ as $(-1)$ raised to the power $0+1+2+\cdots+(n-1)=n(n-1)/2.$

Answer 3

Note that swapping two rows of a matrix changes the sign of its determinant. Given the antidiagonal 1s matrix, how many swaps does it take to turn it into an identity matrix?