Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What suffices to show that two Linear Spaces are Isomorphic? The idea is intuitive though, but I'm not sure how to write it. Assuming exposure to only undergraduate linear algebra.

Say for example, set of all Row vectors in $ \mathbb{R}^n $ and set of all polynomials of degree $ < n$.

share|improve this question
    
The set of all polynomials of degree less than or equal to $n$ is a vector space of dimension $n+1$. –  JSchlather Nov 13 '12 at 21:45
    
sorry, my bad. its strictly less than.. made the correction –  Aseem Dua Nov 13 '12 at 21:49

2 Answers 2

up vote 3 down vote accepted

Theorem: Two finite-dimensional vector spaces over a field are isomorphic if and only if they have the same dimension. So it is enough that you find a basis for each of your vector spaces, count their elements, and show that the two basis have the same number of elements.

share|improve this answer
    
Thanks a lot for the answer! –  Aseem Dua Nov 14 '12 at 7:19

The answer by Manos works if you know the vecctor spaces are finite-dimensional and you know (or can easily find) the dimensions.

In general, to show $V$ and $W$ are isomorphic, you have to find an invertible linear transformation from $V$ to $W$. That is, you have to find a function $T:V\to W$ such that

  1. $T(au+bv)=aT(u)+bT(v)$ for all scalars $a,b$ and all $u,v$ in $V$, and

  2. $T$ is one-one and onto.

So at the very least, you have to know what one-one means, and what onto means, and how to prove something is one-one, and how to prove something is onto.

share|improve this answer
    
Your answer is useful, but yes I was more concerned about the finite case. Thanks a lot! –  Aseem Dua Nov 14 '12 at 7:18
    
Even in the finite-dimensional case, you have to know what to do in case you don't know the dimensions. –  Gerry Myerson Nov 14 '12 at 10:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.