Show that if all the row sums of a matrix A belong to C (nxm) are

  • Thread starter Thread starter bernoli123
  • Start date Start date
  • Tags Tags
    Matrix Row Sums
bernoli123
Messages
11
Reaction score
0
show that if all the row sums of a matrix A belong to C (nxm) are zeroes, then A is singular.
Hint. Observe that Ax=0 for x=[1 1 ...1]T
 
Physics news on Phys.org


A singular matrix is one whose rows are linearly dependent. A Gauss-Jordan elimination on a singular matrix will leave you with one or more lines full of zeroes.

Another way to look at it: suppose your matrix A is a representation of some linear system of equations (each row would correspond to one equation). If you have less equations than unknowns, you can solve by choosing arbitrary values for these "orphan" unknowns. Meaning you have infinite solutions (or no solutions at all).

There's many different ways to prove that a matrix is singular given those conditions. One of them could be assuming that A in NOT singular and then testing for linear independence. Another way could be using determinants: you could factor out a 0 and cancel every cofactor in the expansion.

If the general form nxm gives you trouble, try working it out for a 2x2 and then expand on that.
 


thank you very much
 
##\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