Is the matrix exponential for a given matrix $\mathbf{A}$ always the same?

Question

If I was given a matrix $\mathbf{A}$ and said to obtain the matrix exponential $$e^{\mathbf{A}t}=\Phi(t)\Phi(0)^{-1}$$ does this matrix $\mathbf{A}$ has only one matrix exponential? Or does it depend on the solution vectors obtained?

I am asking this, because I was solving a problem and the exponential matrix I obtained was different than the one in the instructor's solutions manual. I don't know if this question has been asked here before, I tried looking it up but didn't find anything.

For example, I was asked to obtain the matrix exponential for $$\mathbf{A}=\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}$$

the answer I obtained was $$e^{t}\begin{bmatrix} 1 & 2t & 4t^{2} +5t \\ 0 & 1 & 4t \\ 0 & 0 & 1 \end{bmatrix}$$

however, the one in the solutions manual was $$e^{t}\begin{bmatrix} 1 & 2t & 4t^{2} +3t \\ 0 & 1 & 4t \\ 0 & 0 & 1 \end{bmatrix}$$

Answer 1

$\exp tA$ is a well-defined function of $tA$. It only depends on $tA$.

The Jordan Decomposition is $$ \begin{bmatrix} 1&2&3\\0&1&4\\0&0&1\end{bmatrix} =\begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix} \begin{bmatrix} 1&1&0\\0&1&1\\0&0&1\end{bmatrix} \begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix}^{-1} $$ Next $$ \begin{bmatrix} 1&1&0\\0&1&1\\0&0&1\end{bmatrix}^n =\begin{bmatrix} 1&\binom{n}{1}&\binom{n}{2}\\0&1&\binom{n}{1}\\0&0&1\end{bmatrix} $$ Therefore, $$ \exp t\begin{bmatrix} 1&1&0\\0&1&1\\0&0&1\end{bmatrix} =e^t\begin{bmatrix} 1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix} $$ Thus, $$ \begin{align} \exp t\begin{bmatrix} 1&2&3\\0&1&4\\0&0&1\end{bmatrix} &=e^t\begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix} \begin{bmatrix} 1&t&\frac{t^2}2\\0&1&t\\0&0&1\end{bmatrix} \begin{bmatrix} 1&0&0\\0&\frac12&-\frac3{16}\\0&0&\frac18 \end{bmatrix}^{-1}\\[6pt] &=e^t\begin{bmatrix} 1&2t&3t+4t^2\\0&1&4t\\0&0&1\end{bmatrix} \end{align} $$ So my answer agrees with the book.

Without seeing your work, I cannot comment on where the problem might be.

Answer 2

Although there are many ways to compute it, the exponential of a matrix is unique. If you’re solving a differential equation, though, you might come up with a different partition of the arbitrary constants of integration from the book solution. In this case, however, it looks like you’ve made an error in computing this matrix exponential.

No doubt you’ve already found that the only eigenvalue has geometric multiplicity one, so let $N=A-I$. The matrices $N$ and $I$ commute, therefore $e^{tA}=e^{t(I+N)}=e^te^{tN}$. In addition, $N^3=0$, therefore $$e^{tA}=e^t\left(I+tN+\frac{t^2}2N^2\right).$$ Without computing $N^2$ you can see from this that the element in the upper-right corner must be of the form $3t+ct^2$, so your calculation went awry somewhere.

Answer 3

Yes, the matrix exponential is uniquely defined by the exponential series. As every matrix commutes with the identity, you get $$ \exp((I+B)t)=e^t\exp(Bt)=e^t(I+Bt+\frac12B^2t^2+\frac16B^3t^3+...) $$ Here $B=\pmatrix{0&2&3\\0&0&4\\0&0&0}$ is nilpotent, $B^3=0$, the same for all higher powers, so that the exponential series only consists of the first three terms, $$ \exp(At)=e^t\left(\pmatrix{1&0&0\\0&1&0\\0&0&1}+\pmatrix{0&2&3\\0&0&4\\0&0&0}t+\frac12\pmatrix{0&0&8\\0&0&0\\0&0&0}t^2\right) =e^t\pmatrix{1&2t&3t+4t^2\\0&1&4t\\0&0&1} $$