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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.

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    $\begingroup$ That's just the definition of matrix exponentials: For $M\in K^{n\times n}$, we have $$\exp(M)=\sum_{k=0}^\infty\frac{M^k}{k!}$$ $\endgroup$ Jul 21 at 12:54
  • $\begingroup$ @Shaun Is there some other definition of $\exp(M)$ (for a matrix $M$) that you are more familiar with? $\endgroup$ Jul 21 at 13:09
  • $\begingroup$ Welcome to Mathematics Stack Exchange. Are you familiar with the Taylor series for the exponential function? $\endgroup$ Jul 21 at 13:14
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    $\begingroup$ @J.W.Tanner Ah now I remember I learned this before. Sorry for asking such a stupid question. I major in chemistry so I'm not so familiar with this. $\endgroup$
    – Shaun Han
    Jul 21 at 13:22
  • $\begingroup$ You can use other definitions for $e^M$. For example, solution of the differential equation $$\frac{d}{dx} F(x) = M F(x)$$ is $F(x) = e^{xM}F(0)$. $\endgroup$
    – GEdgar
    Jul 21 at 14:26
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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!

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