Don't understand this simple vector space problem

octol
Messages
61
Reaction score
0
Don't understand this reasoning with respect to linear operators.

Let S and T be linear operators on the finite dimensional vector space V. Then assuming the composition ST is invertible, we get
\text{null} \; S \subset \text{null} \; ST

Why is that? I thought hard about it but I simply cannot follow. Is it not possible to have an element x of V that is in the nullspace of S but not in the nullspace of ST ? i.e. S maps x to 0 but T maps x to y where S don't map y to 0 ?
 
Physics news on Phys.org
If ST is invertible, then the null space of it is zero. I think you ought to re-examine your question.

It is certainly true that null(T) is a subspace of null(ST).
 
yes I understand that the null space of ST i zero, and that null(T) is a subspace of null(ST), but how can we say that null(S) is a subspace of null(ST) ? I must be doing some kind of faulty thinking :(
 
You can't say null(S) is a subspace of null(ST), in general. It isn't. Trivially. Howver, you asserted that ST was invertible, and at no point attempted to use this fact. Thus null(ST)=0, so you're asking 'is null(S) a subspace of the zero vector space'. Well, what is the only subspace of the zero space? I.e. is S injective?
 
it depends which convention on composition is being used. some people write ST for first S then T, but not me.
 
##\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