It is currently Mon Oct 23, 2017 12:45 pm


All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: Determinant
PostPosted: Thu Jun 09, 2016 12:04 pm 
Administrator
Administrator
User avatar

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 803
Location: Larisa
Let \( A\in \mathcal{M}_n\left ( \mathbb{C} \right ) \) with \( n\geq 2 \). If \( \det \left ( A+X \right )=\det A+\det X \) for every matrix \( X \in \mathcal{M}_n\left ( \mathbb{C} \right ) \) , then prove that \( A=\mathbb{O} \).

_________________
Imagination is much more important than knowledge.
Image


Top
Offline Profile  
Reply with quote  

 Post subject: Re: Determinant
PostPosted: Thu Jun 09, 2016 12:05 pm 

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
Suppose that \(A \neq 0\), say \(A_{ij} \neq 0\) for some \(i,j\). Let \(P\) be any permutation matrix with \(P_{ij}=1\) and let \(Q\) be the matrix obtained from \(P\) by changing its \(ij\)-entry to \(0\). Finally let \(X = xQ\) where \(x \in \mathbb{C}\).

We have that \(\det(X) = 0\) and that \(\det(X) = \det(A+X) - \det(A)\) is a polynomial in \(x\). Furthermore, the coefficient of \(x^{n-1}\) of this polynomial is \(\pm A_{ij}\) depending on the sign of the corresponding permutation. So the polynomial is not identically zero, a contradiction.


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net