What is the Algebraic Proof of Linear Transformation Composition with Addition?

autre
Messages
116
Reaction score
0

Homework Statement



Prove:

Let V be a vector space over the field F . If A,B,C\in L(V) , then A\circ(B+C)=A\circ B+A\circ C .

The Attempt at a Solution



Note that A\circ B\in L(V) means A\circ B(\mathbf{v})=A(B(\mathbf{v})). Suppose (\alpha_{jk})_{j,k=1}^{n} and (\beta_{jk})_{j,k=1}^{n} are matrices of A and B and (\gamma_{jk})_{j,k=1}^{n} is a matrix of C . Then, B+C=(\beta_{jk}+\gamma_{jk})_{j,k=1}^{n} and A\circ(B+C)=A((B+C))=\sum_{i=1}^{n}\alpha_{ji}(\beta_{ik}+\gamma_{ik})...

I'm a little stuck at this point. Any ideas?
 
Physics news on Phys.org
you just need to continue the algebra a little further...

\sum_i \alpha_{ji}(\beta_{ik} + \gamma_{ik}) = \left(\sum_i\alpha_{ji}\beta_{ik}\right) + \left(\sum_i\alpha_{ji}\gamma_{ik}\right) = \dots
 
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