Why is $\sum_k \frac{\lambda^k}{k!} A^k = $ exp$(\lambda A)$? I am reading a paper about simililarity between graphs in this paper. In page 10, there is this equation for a diffusion kernel
$k(i,j) = [\sum_k \frac{\lambda^k}{k!} A^k]_{ij} = $ exp$(\lambda A)]_{ij}$
where $A$ is a square matrix with all values between 0 and 1, $\lambda$ is a value between 0 and 1, $k$ is a positive integer.
I don't quite understand how exp$(\lambda A)$ is derived. Can someone explain it to me? Thanks in advance.
 A: By definition, the Taylor Series of $$e^x=\sum_{n=0}^\infty\frac{f^{(n)}(a)(x-a)^n}{n!}$$ as the coefficients of $(x-a)^n$ are just the pattern of the nth derivative of $$f(x)=\sum_{n=0}^\infty c_n(x-a)^n$$ as seen in this Paul’s notes article where $f^{(n)}$ is the nth derivative where the summation form above is assumed.
Derivation:
$$f^{(0)}(a)=f(a)\mathop=^{\text{def}}c_0,f^{(1)}(x)=c_1+2c_2(x-a)^1+…=c_1,f^{(2)}(x)=2c_2+2(1)(x-a)^0+… $$
Continue the process, plug in x=a by definition so that the (x-a) parts are 0, and solve for $c_k$. Then, you will get the familiar Taylor series pattern.
Finding the nth derivative of $e^{x-a}$ where the approximation will start to look like $e^x$ at x=a. So we can just substitute x-a$\to$ x. Therefore we can just assume a=$0$ and not worry about substituting other values of a into the nth derivative for a more complicated sum expression.
$$f^{(0)}(0)=e^0=1, f^{(1)}(x)=e^0=1\implies f^{(n)}(x)=1\implies e^x=\sum_{n=0}^\infty\frac{(x-a)^n}{n!}\mathop=^{a=0}= \sum_{n=0}^\infty\frac{x^n}{n!}\mathop=^{x=\lambda A}= \sum_{n=0}^\infty\frac{\lambda^n}{n!}A^n $$
If we assume x is a matrix, it works the same way. Finally, here is a graph of the expansion. Please correct me and give me feedback!
