Is Det(A+B) Equal to Det(A) + Det(B)?

  • Thread starter Thread starter phyin
  • Start date Start date
phyin
Messages
7
Reaction score
0
det(A+B) ?? det(A) + det(B)

I'm guessing greater than but I'm not too sure. I need a proof on this so I can be assured of it and then use the statement to prove something else.

any hint (or link to proof) would be much appreciated.

edit:

x*det(A) ?? det(x*A)
what's the relation there?
 
Physics news on Phys.org


Neither holds, take

A=B=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)

for one equality and


A=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right),~B=-A

for the other.

For the last question, we do have the equality

det(xA)=x^n detA

where A is the rank of the matrix. From this we can deduce that neither of the inequalities between det(xA) and xdet(A) holds true. Just take x>1 and x<1.
 


thanks. i better find an alternative way to my proof ;s
 


Another simple counter-example is
A = \begin{pmatrix} a &amp; 0 \\ 0 &amp; 0 \end{pmatrix} \qquad <br /> B = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; b \end{pmatrix}

det(A) = det(B) = 0, det(A+B) = ab
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top