0
$\begingroup$

How to prove the determinant of

$$A=(a_{ij})_{n\times n}=b_i^2-b_j^2$$ where $b_1,b_2,...,b_n$ are some distinct real numbers , $n\geq 3$ and $i,j=1,2,...,n$ is zero?

Here, $A$ is skew symmetric and I know odd order skew symmetric matrix has determinant zero. How about the even case? or Is there any other method to show $det(A)=0$ ?

$\endgroup$
  • $\begingroup$ It is zero when $n \geq 3$. Indeed, this is a particular case of the fact that $\det\left(\left(x_i+y_j\right)_{1\leq i\leq n,\ 1\leq j\leq n}\right) = 0$ whenever $n \geq 3$. For this fact, see, e.g., Example 5.28 in Notes on the combinatorial fundamentals of algebra, 7 November 2017. $\endgroup$ – darij grinberg Nov 10 '17 at 3:44
  • 3
    $\begingroup$ The comment by @darijgrinberg gives a very nice explanation, and it is impressive to see this writing which is nearly 800 pages long! (I wish that I could study this some day!) Nevertheless a very quick explanation is possible, you do not quite have to find an example in the middle of $>780$ pages. The matrix $(b_i^2)$ has rank $1$ because each row is constant. The matrix $(b_j^2)$ has rank $1$ because each column is constant. The difference is a matrix of rank $2$ at most. If $n \geq 3$, an $n \times n$ matrix of rank $2$ or less has determinant zero. $\endgroup$ – Zach Teitler Nov 10 '17 at 5:45
  • 1
    $\begingroup$ @ZachTeitler: Thanks, but there's probably little new in there for you :) I was focussing on writing up in full detail the standard arguments that often end up in the no-mans-land between linear algebra and combinatorics. $\endgroup$ – darij grinberg Nov 10 '17 at 5:54
  • $\begingroup$ @darijgrinberg Well, most of the main topics are familiar, but I'm sure that I could learn a lot from the examples. Hopefully some day. (Or perhaps I will convince a student to read it and tell me about it. :-) ) $\endgroup$ – Zach Teitler Nov 10 '17 at 6:32
2
$\begingroup$

From the comments: $A$ is the difference of two rank-one matrices, so that the rank of $A$ is at most two. It follows that $\det A = 0$ if $n \ge 3$.

$\endgroup$
1
$\begingroup$

You can't do it, at least not for all $n$, as the question was originally asked.

Pick $b_1 = 1$ and $b_2 = 0$ to get the matrix $$ \pmatrix{0 & 1 \\ -1 & 0} $$ whose determinant is not zero.

$\endgroup$
  • 1
    $\begingroup$ Fine! Actually we consider the case $n \geq3$ (I have edited the question.) $\endgroup$ – user444830 Nov 10 '17 at 5:28
  • 1
    $\begingroup$ Actually, you asked about the general case. Changing the question after folks have answered what you asked originally is a bit rude. Better, in the future, to make sure you're asking the question you wanted to ask right from the start. $\endgroup$ – John Hughes Nov 10 '17 at 11:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy