Linear Operators and Matrices

1. Oct 24, 2014

albega

Suppose we have linear operators A' and B'. We define their sum C'=A'+B' such that
C'|v>=(A'+B')|v>=A'|v>+B'|v>.

Now we can represent A',B',C' by matrices A,B,C respectively. I have a question about proving that if C'=A'+B', C=A+B holds. The proof is

Using the above with Einstein summation convention,
C|v>=A|v>+B|v>
and so component i on each side matches. Then
Cijvj=Aijvj+Bijvj
which holds for any |v>, so
C=A+B
as this is how we define matrix addition.

However, why couldn't I have written
Cijvj=Aikvk+Bilvl
because I have changed only dummy variables, not affecting the sum. This would then not lead to Cijvj=Aijvj+Bijvj. I'm assuming it's something to do with the fact the next step sort of stops this sum from happening anyway, but I'm not sure.

2. Oct 24, 2014

Orodruin

Staff Emeritus
Yes, you could have written that, but it would change nothing since the only difference is in what you call the dummy variables. You could just rename them back and write it back on your original form.

3. Oct 24, 2014

albega

It's just that if I do say
Cijvj=Aikvk+Bilvl
and then I write
Cij=Aik+Bil
It looks like a different story, and we can't see that the k and l were initially dummy variables and so we can't simply say it's ok to change them both to j.

Ahh, actually, if we do that then we can't cancel out the v components anymore from each of the sums can we, which answers my question...

4. Oct 24, 2014

Orodruin

Staff Emeritus
k and l are dummy variables, you cannot just take them away from the v, you would end up with an expression that does not make sense (you cannot have different free indices in Cij). However, you can rename them both to j (they are dummy indices after all) and then factorise the vj out.

5. Oct 26, 2014

Fredrik

Staff Emeritus
Use the definition of matrix multiplication: $(AB)_{ij}=A_{ik}B_{kj}$.
\begin{align}
&C_{ij}v_j=(Cv)_i\\
&A_{ik}v_k+B_{il}v_l=(Av)_i+(Bv)_i=(Av+Bv)_i
\end{align} Since the left-hand sides are equal, the right-hand sides are equal, and we have $Cv=Av+Bv=(A+B)v$. If this holds for all v, then $C=A+B$.