A Question in Linear Tranformations

  • Thread starter Thread starter beanz
  • Start date Start date
  • Tags Tags
    Linear
beanz
Messages
4
Reaction score
0
hi

I am a college student who has just started doing Linear Algebra. This is not a homework question, just something abot Linear Transformations that i don't understand. I hope someone can help me. Here it goes:

Consider Vector spaces V and W (over R^4) and a matrix 'A' which maps an element from V to W.

1. Why is it that the basis for the column space for A is exactly the basis for the range space of V in W? [i.e why is dim(col(A))=dim (range(V))]

2. Why is dim (V) = dim (Null space of V) + dim (range space of V in W)?

Thanx in advance :)
 
Physics news on Phys.org
The basis for V is, of course, (1, 0, 0, 0), (0, 1, 0,0), (0, 0, 1, 0), (0, 0, 0, 1). Write those as columns and multiply each by A. Do you see that multiplying A times (1, 0, 0, 0), you get exactly the first column of A? And that multiplying A time (0, 1, 0, 0) gives exactly the second column of A? Write any vector in V as a linear combination of (1, 0, 0, 0), etc. and multiplying by A gives the result, in W, as a linear combination of the columns of A.
 
Hahaha, Thanks. It is so obvious now that you mention it.
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
##\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...
Back
Top