Matrix exponential of non diagonalizable matrix?

Solution 1:

There are two facts that are usually used for this computation:

Theorem: Suppose that $A$ and $B$ commute (i.e. $AB = BA$). Then $\exp(A + B) = \exp(A)\exp(B)$

Theorem: Any (square) matrix $A$ can be written as $A = D + N$ where $D$ and $N$ are such that $D$ is diagonalizable, $N$ is nilpotent, and $ND = DN$

With that, we have enough information to compute the exponential of every matrix.

For your example, we have $$ D = \pmatrix{1&0\\0&1} = I, \quad N = \pmatrix{0&0\\1&0} $$ we find that $$ \exp(D) = eI\\ \exp(N) = I + N + \frac 12 N^2 + \cdots = I + N + 0 = I + N $$ So, we have $$ \exp(D + N) = \exp(D) \exp(N) = (eI)(I+N) = e(I+N) = \\ \pmatrix{e&0\\e&e} $$

Solution 2:

In the $2\times2$ case, you can find the exponential of a matrix $A$ without having to decompose it into $BMB^{-1}$ form. In particular, you only need the eigenvalues—you don’t need to find any eigenvectors. There are three cases, as follows.

Distinct Real Eigenvalues: Let $P_1 = (A-\lambda_2I)/(\lambda_1-\lambda_2)$ and $P_2 = (A-\lambda_1I)/(\lambda_2-\lambda_1)$, where $\lambda_1,\lambda_2$ are the eigenvalues. These two matrices are projections onto the eigenspaces corresponding to $\lambda_1$ and $\lambda_2$, respectively. They have some handy properties: $P_1P_2=P_2P_1=0$, $P_1+P_2=I$, and $A=\lambda_1P_1+\lambda_2P_2$. Using these projections, $$\exp(tA)=\exp(t\lambda_1P_1+t\lambda_2P_2)=\mathrm e^{\lambda_1t}P_1+\mathrm e^{\lambda_2t}P_2.$$

Repeated Eigenvalue: Let $G=A-\lambda I$, where $\lambda$ is the eigenvalue. By the Cayley-Hamilton theorem, $(A-\lambda I)^2=0$, so $G$ is nilpotent. Since $G$ and $\lambda I$ commute, $\exp(tA)=\exp(\lambda tI)\exp(tG)$, but $\exp(tG)=I+tG$ (expand using the power series), so $$\exp(tA) = \mathrm e^{\lambda t}(I+tG).$$

Complex Eigenvalues: The eigenvalues are of the form $\lambda=\alpha\pm\mathrm i\beta$, and the characteristic equation is $(\lambda-\alpha)^2+\beta^2=0$. By the Cayley-Hamilton theorem, $(A-\alpha I)^2+\beta^2I=0$, so the traceless matrix $G=A-\alpha I$ satisfies $G^2=-\beta^2 I$. Writing $A=\alpha I+G$, we have $\exp(tA)=\exp(\alpha I)\exp(tG)$. Expanding the latter as a power series and using the above equality, $$ \exp(tG) = I+tG-\frac{\beta^2t^2}{2!}I-\frac{\beta^2t^3}{3!}G+\frac{\beta^4t^4}{4!}I+\dots. $$ The coefficient of $I$ is the power series for $\cos{\beta t}$, while the coefficient of $G$ is the power series for $(\sin{\beta t})/\beta$. Thus, $$ \exp(tG) = \cos{\beta t}\;I+[(\sin{\beta t})/\beta]\;G \\ \text{and} \\ \exp(tA) = \mathrm e^{\alpha t}\exp(tG). $$