Definition of Kronecker Delta

Question

Is $\delta _{mn}=1$ when $m\neq n$, and $\delta _{mm}=0$?

I am not very good at Math. So would you give me the answer and explanation please?

Answer 1

No, the Kronecker delta is defined as $\delta_{mn}=1$ when $m=n$ and $\delta_{mn}=0$ when $m\ne n$. (As Berrick Fillmore said).

It is useful in certain matrices and sum. For example, the identity matrix $I$ has entries $(\delta_{ij} : 1\le i,j\le n)$. And $$\sum_{i=1}^n a_i \delta_{ij} = a_j$$

Answer 2

Late to the party, but here's another way to see why this quantity is special: let $V$ be a vector space and $V^*$ its dual. Define a tensor $\delta\colon V^* \times V \to \Bbb R$ by $\delta(f,v) = f(v)$. If $\mathcal{B} = (e_1,\ldots,e_n)$ and $\mathcal{B}^* = (e^1,\ldots,e^n)$ is the dual basis, the components of $\delta$ are, by definition: $$\delta^i_{\;j}\doteq \delta(e^i,e_j) = e^i(e_j) = \begin{cases} 1, & \mbox{if }i=j \\ 0, & \mbox{if } i \neq j\end{cases}$$ Fun thing is: the numerical value of the components does not depend on the bases $\mathcal{B}$ and $\mathcal{B}^*$ themselves. Also, if $V$ has a inner product $\langle\cdot,\cdot\rangle$, you can raise and lower indexes of $\delta$. Using orthonormal bases, one concludes that the numerical values of $\delta^i_{\;j}$, $\delta_i^{\;j}$, $\delta_{ij}$ and $\delta^{ij}$ are all the same.