Understanding Singular Linear Maps: R^m -> R^n

annoymage
Messages
360
Reaction score
0

Homework Statement



the question here said

is L, linear transformation/mapping is singular?

i'm still googling the definition singular linear map,

can anyone give me the definition please T_T

p/s; i thought it L maybe the matrix representation, but the question

L : R^m -> R^n

but aren't matrix representation are define on L: M(m,1) -> M(n,1)

should i take transpose of R^m and R^n ?
 
Last edited:
Physics news on Phys.org
If L is a linear map between finite-dimensional vector spaces (such as \mathbb{R}^m and \mathbb{R}^n), then it can be represented by a matrix. In fact, it can be represented by many different matrices, each corresponding to different choices of basis. But sometimes the answer to a question can be clearer if you don't focus on the matrix representation. Unfortunately, you did not fully state the question! (i.e., how is L defined?)

In any case, a singular linear map is simply one that does not have an inverse. It can fail to have an inverse in one (or both) of two ways: (1) it is not injective, meaning there exist a \neq b \in \mathbb{R}^m such that L(a) = L(b); or (2) it is not surjective, meaning that there exists some b \in \mathbb{R}^n such that L(a) \neq b for all a \in \mathbb{R}^m.

If m \neq n, then L is guaranteed to be singular. (Why?)
 
thanks for the definition.

And your question, I can see If m<n then not surjective means singular

if m>n then not injective means singular. But still working on that. Tomorrow i'll try to post, help check it.

Anyway, i don't understand this,
jbunniii said:
If L is a linear map between finite-dimensional vector spaces (such as \mathbb{R}^m and \mathbb{R}^n), then it can be represented by a matrix. In fact, it can be represented by many different matrices, each corresponding to different choices of basis.

what topic should i study for this thing?
 
annoymage said:
And your question, I can see If m<n then not surjective means singular

Correct, surjectivity is impossible when m &lt; n, and injectivity is impossible when m &gt; n.
Anyway, i don't understand this

what topic should i study for this thing?

Well, the topic is linear algebra, but you can study it at several levels of sophistication. Are you taking a course in it? If so, have you seen the definition of a linear map between two vector spaces? It should look something like this:

If V and W are vector spaces over the same field K and L : V \rightarrow W, then L is a linear map if the following are true for all v, v_1, v_2 \in V, and k \in K:

(1) L(v_1 + v_2) = L(v_1) + L(v_2)

(2) L(kv) = kL(v)
 
annoymage said:
what topic should i study for this thing?

doesn't matter, i think i found it, i'll read it forthwith
 
jbunniii said:
Well, the topic is linear algebra, but you can study it at several levels of sophistication. Are you taking a course in it? If so, have you seen the definition of a linear map between two vector spaces? It should look something like this:

If V and W are vector spaces over the same field K and L : V \rightarrow W, then L is a linear map if the following are true for all v, v_1, v_2 \in V, and k \in K:

(1) L(v_1 + v_2) = L(v_1) + L(v_2)

(2) L(kv) = kL(v)

yea i did pass that topic, only my lecture note doesn't define what singularity of linear map is, anyway thank you so much
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top