function of matrix and eigen values

Question

I want to calculate exp(A), A is matrix, with numann series. is this series depend of matrix's eigen values? for example if it's eigen values are large, is numann series useful for this function?

Answer 1

If we have a diagonalizable matrix $\mathbf{A}=\mathbf{P}\mathbf{\Lambda}\mathbf{P}^{-1}$, where $\mathbf{\Lambda}=\operatorname{diag}(\lambda_{1},\dots,\lambda_{n})$, and $\lambda_{i}$ are the eigenvalues of $\mathbf{A}$, and we wish to calculate $\exp(\mathbf{A})$, we use the Taylor series for $\exp(x)$:

$$\exp(x)=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$$

We can also note that $\forall m\in\mathbb{N}$, we have: $(\mathbf{PM}\mathbf{P}^{-1})^{m}=\mathbf{P}\mathbf{M}^{m}\mathbf{P}^{-1}$, which can be shown by writing out the multiplication explicitly and using the identity $\mathbf{P}^{-1}\mathbf{P}\equiv \mathbf{I}_{n\times n}$. Therefore we have $\exp(\mathbf{A})=\mathbf{P}\exp(\mathbf{\Lambda})\mathbf{P}^{-1}$:

$$\exp(\mathbf{A})=\mathbf{P}\left(\sum_{k=0}^{\infty}\frac{\mathbf{\Lambda}^{k}}{k!}\right)\mathbf{P}^{-1}=\mathbf{P}\left(\sum_{k=0}^{\infty}\frac{1}{k!}\operatorname{diag}(\lambda_{1}^{k},\dots,\lambda_{n}^{k})\right)\mathbf{P}^{-1}$$

Which using the standard rules of scalar-matrix multiplication gives us:

$$\exp(\mathbf{A})=\mathbf{P}\operatorname{diag}\left(\sum_{k=0}^{\infty}\frac{\lambda_{1}^{k}}{k!},\dots,\sum_{k=0}^{\infty}\frac{\lambda_{n}^{k}}{k!}\right)\mathbf{P}^{-1}=\mathbf{P}\operatorname{diag}(\exp(\lambda_{1}),\dots,\exp(\lambda_{n}))\mathbf{P}^{-1}$$