How do I handle Kronecker Delta in my homework?

[Bellin12]

Homework Statement

I am a bit confused with how to deal with Kronecker Delta.
I need to show that P,i = ( P*δij ),j

The i and j are subscripts.

Homework Equations

The Attempt at a Solution

P,i = ( P*δij ),j = P,j*δij + P*δij,j
I assumed I could get rid of P*δij,j leaving me with P,j*δij.
After this I am stuck. I'm not sure if I am supposed to consider Kronecker Delta as the Identity Matrix now or not. If I do, I end up with P,j which technically would equal P,i.

 

[vela]

What exactly is P here?

 

[quietrain]

is it like this?

we have
{P * d_ij }_j

since d_ij is 0 for all i,j except i=j where it is 1,

then {P * d_ij }_j
= {P * 1 }_i ==> since i = j
= P_ii was taught to think of the kronecker delta as a sieving function, i.e, it 'sieves' out the index 'j' and replace it by 'i' , as per above

 

[vela]

The Attempt at a Solution

P,i = ( P*δij ),j = P,j*δij + P*δij,j
I assumed I could get rid of P*δij,j leaving me with P,j*δij.
After this I am stuck. I'm not sure if I am supposed to consider Kronecker Delta as the Identity Matrix now or not. If I do, I end up with P,j which technically would equal P,i.

There's no need to assume the second term is 0. δij is equal to 0 or 1, so if you differentiate it, you get 0.

The second term isn't equal to P_,j. The index j is a dummy index you're summing over. Just expand the summation out to see what's happening:
$$P_{,j}\delta_{ij} = P_{,1}\delta_{i1} + P_{,2}\delta_{i2} + \cdots + P_{,n}\delta_{in}$$In only one term, the ith term, does the Kronecker delta not vanish, so you end up with P_,i.