0
$\begingroup$

Let $E=\begin{pmatrix} (1-n)e_1 & e_2 & ... & e_n\\ e_1 & (1-n)e_2 & ... &e_n\\ . & . & . &.\\ . & . & . &.\\ e_1 & e_2 & ... & (1-n)e_n \end{pmatrix}$ an $n \times n$ matrix. I need to prove that the determinant is zero. I thought about using guassian elimination, and somehow $e_{nn}$ (position) will lead to zero. So, as a triangular matrix with a zero in the diagonal means $Det(E)=0$, then it is solved.

However, I would like to know if there's a better way to prove it, or a hint, and if what I thought is correct.

$\endgroup$
3
$\begingroup$

HINT: If you sum all the rows, which is a non-trivial linear combination of them, you get... And this would imply the determinant is zero because any non-trivial linear combination of the rows of an invertible (i.e. non-zero determinant) matrix cannot be this value.

$\endgroup$
1
  • $\begingroup$ Excelent approach, I understand it, thank you a lot! $\endgroup$ Nov 6 '20 at 17:38
1
$\begingroup$

First $(1)$, factor out the coefficients. Second $(2)$, add all rows to the first row: $$\begin{vmatrix} (1-n)e_1 & e_2 & ... & e_n\\ e_1 & (1-n)e_2 & ... &e_n\\ . & . & . &.\\ . & . & . &.\\ e_1 & e_2 & ... & (1-n)e_n \end{vmatrix} $$$$ \overset 1=(e_1\cdots e_n)\begin{vmatrix} (1-n) & 1 & ... & 1\\ 1 & (1-n) & ... &1\\ . & . & . &.\\ . & . & . &.\\ 1 & 1 & ... & (1-n) \end{vmatrix} $$$$\overset 2 =(e_1\cdots e_n)\begin{vmatrix} (1-n)+(n-1) & (1-n)+(n-1) & ... & (1-n)+(n-1)\\ 1 & (1-n) & ... &1\\ . & . & . &.\\ . & . & . &.\\ 1 & 1 & ... & (1-n) \end{vmatrix} \\= (e_1\cdots e_n)\begin{vmatrix} 0 & 0 & ... & 0\\ 1 & (1-n) & ... &1\\ . & . & . &.\\ . & . & . &.\\ 1 & 1 & ... & (1-n) \end{vmatrix}$$$$=0$$

$\endgroup$
1
  • 1
    $\begingroup$ Nice way to write $(e_1,..,e_n)$, it makes it easier, thanks. $\endgroup$ Nov 6 '20 at 17:38
1
$\begingroup$

Consider the vector $v = [1,1,\ldots, 1]^T$. You can calculate component wise that $E^Tv = 0$ which implies that $E$ is singular so $\det(E^T) = 0$ and since $\det(E)=\det(E^T)$ we have that $$ \det(E)=\det(E^T) = 0$$ as desired.

$\endgroup$
1
  • $\begingroup$ Great! I appreciate the answer. $\endgroup$ Nov 6 '20 at 17:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.