How can one determine $e^{tA}$, where $A =\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix}$?

Question

Let $A =\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix}$ be given.

How can one compute the matrix function $e^{tA}$? Do you have to decompose $A$, if so how?

Answer 1

I'll make use of the following theorem, which gives you a general method for computing matrix exponentials, even when the matrix is not diagonalizable.

Theorem. Let $\lambda_1,\dots,\lambda_m$ denote the different eigenvalues of the matrix $A\in\mathcal{M}_{n\times n}(\mathbb{C})$, let $n_1,\dots,n_m$ denote their respective (algebraic) multiplicities, and let $f$ be an entire function. Then $$f(A)=g(A),$$ where $g$ is the polynomial with $\deg(g)\leq n-1$ uniquely determined by the conditions $$g^{(j)}(\lambda_k)=f^{(j)}(\lambda_k)$$ for all $k\in\{1,\dots,m\}$ and $j\in\{0,\dots,n_k-1\}$.

While this might be a bit overkill for this problem, it still illustrates a useful general technique. So we start by finding the eigenvalues of $A$ and their multiplicities. It's easy to check that

$$\det(A-\lambda I)=(\lambda-1)(\lambda-2),$$

so the eigenvalues are $1$ and $2$, both with multiplicity $1$. Now we wish to compute $e^{tA}$, so if we fix $t$, then our entire function is given by $f(z)=e^{tz}$. We thus seek a polynomial $g$ of degree at most $1$ with

$$g(1)=e^{t}\quad\text{and}\quad g(2)=e^{2t}.$$

Being a bit clever we can write

$$g(z)=p_1(z)e^t+p_2(z)e^{2t}$$

for some polynomials $p_1,p_2$ of degree at most $1$ with

$$p_1(1)=1,\quad p_2(1)=0,$$ $$p_1(2)=0,\quad p_2(2)=1.$$

From this it's easy to immediately see that

$$p_1(z)=-(z-2)\quad\text{and}\quad p_2(z)=z-1,$$

and so

$$g(z)=-(z-2)e^t+(z-1)e^{2t}.$$

It now follows that

$$e^{tA}=g(A)=-(A-2I)e^t+(A-I)e^{2t}=\begin{pmatrix}1&0\\-1&0\end{pmatrix}e^t+\begin{pmatrix}0&0\\1&1\end{pmatrix}e^{2t}=\begin{pmatrix}e^t&0\\e^{t}(e^t-1)&e^{2t}\end{pmatrix}.$$

Of course there are other methods that can be used as well. For example if $A$ is diagonalizable, we can write $A=TDT^{-1}$ and compute it as $e^{tA}=Te^{tD}T^{-1}$, where $e^{tD}$ is computed by just applying the exponential function to the diagonal elements. One could also do what is called an invariant subspace decomposition and write $A$ as a block diagonal matrix and apply the exponential function to each block, and use that, with the right methods, the blocks will be on the form $\lambda_jI+N$ where $\lambda_j$ is an eigenvalue and $N$ is nilpotent. And there are certainly other methods that can be used as well (such as just using the power series definition if your matrix is nice enough). Hopefully this illustrates how matrix exponential can be computed.