# 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.

• That's just the definition of matrix exponentials: For $M\in K^{n\times n}$, we have $$\exp(M)=\sum_{k=0}^\infty\frac{M^k}{k!}$$ Jul 21 at 12:54
• @Shaun Is there some other definition of $\exp(M)$ (for a matrix $M$) that you are more familiar with? Jul 21 at 13:09
• Welcome to Mathematics Stack Exchange. Are you familiar with the Taylor series for the exponential function? Jul 21 at 13:14
• @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. Jul 21 at 13:22
• You can use other definitions for $e^M$. For example, solution of the differential equation $$\frac{d}{dx} F(x) = M F(x)$$ is $F(x) = e^{xM}F(0)$. Jul 21 at 14:26

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.
$$f^{(0)}(a)=f(a)\mathop=^{\text{def}}c_0,f^{(1)}(x)=c_1+2c_2(x-a)^1+…=c_1,f^{(2)}(x)=2c_2+2(1)(x-a)^0+…$$
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!