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Calculate the determinant of the $2n \times 2n$ matrix with entries equal to zero on the main diagonal, equal to $1$ below and equal to $-1$ above. I'll denote this matrix $A_{2n}$.

So for example you have $A_{2} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} $ and $A_{4}= \begin{bmatrix} 0 & -1 & -1 &-1 \\ 1 & 0 & -1 & -1\\ 1 & 1 & 0 & -1 \\ 1 & 1 & 1 & 0\\ \end{bmatrix} $

From trying it out by hand I think you have $\det(A_{2n})=1$ for all odd $n$ , or $=-1$ for all even $n$.

Can anyone come up with some way to prove this?

From doing these two examples my algorithm seems to be something like this:

  1. Switch the first and second rows (multiplies det by $-1$)
  2. If the matrix is now upper triangular, calculate the determinant by product of main diagonal.
  3. If not, make a new pivot and clear that column below.
  4. Swap the third and fourth rows (multiplies det by $-1$).
  5. Repeat from 2.
Alex Wertheim
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user113867
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5 Answers5

4

Start with $A_{2n}$. If you

  1. subtract $2^{nd}$ column from $1^{st}$ column.
  2. subtract $2^{nd}$ row from $1^{st}$ row.
  3. add $1^{st}$ column to all columns except the first two columns.
  4. add $1^{st}$ row to all rows except the first two rows.

you will turn $A_{2n}$ into block diagonal form with a copy of $A_2$ at the first two rows/columns.

Repeat apply the same procedure to the remaining $2n-2 \times 2n-2$ matrix, one can transform $A_{2n}$ into a block diagonal matrix with $n$ copies of $A_2$. As a result,

$$\det A_{2n} = ( \det A_2 )^n = 1.$$

achille hui
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The determinant is always $+1$.

First note that the inverse of $A_{2n}$ is given by $B_{2n}$, where $$B_{ij} = \begin{cases} 0 & \text{ if }i=j\\ (-1)^{j-i} & \text{ if }j>i\\ (-1)^{i-j+1} & \text{ if }j<i\end{cases}$$ This means that $\det(A) = \det(B) = \pm 1$, since both are integer matrices.

We need to rule out $-1$. This is easy though, since the eigenvalues of an even skew symmetric matrix are purely imaginary, which in turn means that the determinant is positive (it is the square of a the Pfaffian). Hence, the determinant is $+1$.

John Smith
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  • How can you show it's invertible? Also, we haven't covered eigenvalues in our class yet unfortunately, is there any method to demonstrate that part of the proof? – user113867 Jan 03 '14 at 22:08
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    @user113867 I have given you the inverse, just multiply $A_{2n}B_{2n}$ to get the identity matrix. – John Smith Jan 03 '14 at 22:14
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Add $x$ to all the entries of the matrix $A_{2n}$ and denote $C_i$ the $i$-th column of $A_{2n}$ and let $J=\underbrace{(1,\ldots,1)^T}_{2n\;\text{components}}$ hence $$\Delta(x)=\det(C_1+xJ,\cdots ,C_{2n}+xJ)=\det(C_1+xJ,C_2-C_1,\cdots,C_{2n}-C_1)\\ =\det(C_1,C_2-C_1,\cdots,C_{2n}-C_1)+x\det(J,C_2-C_1,\cdots,C_{2n}-C_1)=\alpha+\beta x$$ hence $\Delta(x)$ is a polynomial with degree $1$.

Now it's simple to see (determinant of upper and lower triangle matrix) that $\Delta(1)=\Delta(-1)=1$ hence $$\alpha+\beta=\alpha-\beta=1$$ so $\alpha=1$ and $\beta=0$ and finaly $$\det A_{2n}=\Delta(0)=1$$

2

An elementary method of calculating the determinant through row operations is as follows:

  1. Add row $1$ to rows $3$ through $2n$.
  2. Subtract row $2$ to rows $3$ through $2n$.
  3. Swap rows $1$ and $2$.
  4. Repeat steps (1-3) for the lower right $2(n-1) \times 2(n-1)$ submatrix. Since there are an even number of rows, this process always terminates in an upper triangular matrix whose entries alternate $\{1, -1, \ldots, 1, -1\}$.
  5. The determinant is then $\det A_{2n} = ((-1)^n)^2 = 1$, because there were $n$ row swaps, and there are $n$ $-1$'s on the main diagonal.

Incidentally, if we were to consider an odd number of rows, it is easy to see that the determinant is zero.

heropup
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  • This sounds like best way of going about it, but can you elaborate on steps 1 and 2,and not sure I fully understand – user113867 Jan 03 '14 at 22:52
  • Let $r_j$ be the $j^{\rm th}$ row of $A_{2n}$. For $j = 3, 4, \ldots, 2n$, replace $r_j$ with $r_j + r_1 - r_2$. This is steps 1 and 2 of the above procedure. – heropup Jan 03 '14 at 22:59
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another way would be to evaluate the determinant to the first collumn. From this follows that det(A(2n))=det(A(2n-1))-det(B) + det(B) - det(B) + det(B) - det(B). Because det(B) occurs an odd number of times, it's equal to det(A(2n-1)) - det(B)...

det(A(2n-1)) = 0, because A=-transpose(A) and using determinant properties: det(-transpose(A(2n-1)))= (-1)^(2n-1) * det(transpose(A(2n-1))) = - det(A(2n-1)) = det(A(2n-1))

B is of the form : \begin{bmatrix} -1 & -1 & -1 & -1 & -1\\ 0 & -1 & -1 &-1 & -1 \\ 1 & 0 & -1 & -1 & -1\\ 1 & 1 & 0 & -1 &-1\\ 1 & 1 & 1 & 0 & -1\\ \end{bmatrix} with exactly 2n-1 rows and collumns.

Now if we want det(B) we can remove the (-) of the first row by placing it in front and evaluate the determinant to the first row. Out pops : det(B) = - (det(A(2n-2)) - det(A(2n-2)) + det(A(2n-2)) - det(A(2n-2)) + det(A(2n-2)) ...). But, again because B has an odd number of collumns, it follows that all but one term cancels and we get det(B) = - det(A(2n-2))

Taking this together, we see that det(A(2n)) = det(A(2n-2) and from this follows that it's 1. (look at det (A(2)))

camel
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