How to find the trace of exponential of a matrix

Question

Given $$A=\begin{bmatrix}-3&2\\-1&0\end{bmatrix}$$ how to find $e^{trA}$?

$e^A=pe^Dp^{-1}$ where p is invertible matrix and D is diagonal matrix whose diagonal entries are the eigen value of A. Should i proceed with this concept. Please suggest me. Thank you.

Answer 1

A short route, which avoids unneeded diagonalizations, is to compute every exponential of $A$, that is, to look for functions $x$ and $y$ such that, for every $t$, $$e^{tA}=x(t)A+y(t)I$$ Then, $\mathrm{tr}(A)=-3$ and $\det(A)=2$ hence $$A^2=-3A-2I$$ This implies $$x'(t)A+y'(t)I=(e^{tA})'=Ae^{tA}=A(x(t)A+y(t)I)=x(t)(-3A-2I)+y(t)A$$ that is, $$x'(t)A+y'(t)I=(y(t)-3x(t))A-2x(t)I$$ or, equivalently, $$x'(t)=y(t)-3x(t)\quad y'(t)=-2x(t)$$ This implies that $$x''(t)+3x'(t)+2x(t)=0$$ The roots of the polynomial $t^2+3t+2=(t+2)(t+1)$ are $-1$ and $-2$ hence $$x(t)=ue^{-t}+ve^{-2t}$$ for some given constants $(u,v)$, which implies $$y(t)=x'(t)+3x(t)=2ue^{-t}+ve^{-2t}$$ The initial conditions $x(0)=0$ and $y(0)=1$, when applied to these formulas for $x(0)$ and $y(0)$, yield $u=1$ and $v=-1$, hence, for every $t$, $$e^{tA}=(e^{-t}-e^{-2t})A+(2e^{-t}-e^{-2t})I$$ that is, $$e^{tA}=\begin{pmatrix}-e^{-t}+2e^{-2t} & 2e^{-t}-2e^{-2t}\\-e^{-t}+e^{-2t}&2e^{-t}-e^{-2t}\end{pmatrix}$$ from which the value of the matrix $e^{3A}$ follows.

Answer 2

Use the eigendecomposition of $A$, that is, $A=P\Lambda P^{-1}$ where $\Lambda$ is the diagonal matrix of the eigenvalues $\lambda_1$, $\lambda_2$ of $A$ and $P$ is a matrix whose columns are the orthogonalized eigenvectors of $A$.

Then, $e^A$ is computed as $e^A=Pe^\Lambda P^{-1}$ (mentioned here) where $$e^\Lambda =\begin{bmatrix} e^{\lambda_1} & 0 \\ 0 & e^{\lambda_2} \end{bmatrix}.$$ Since you want to compute $e^{9A}$, note that $9A=P(9\Lambda)P^{-1}$, so $e^{9A}=Pe^{9\Lambda}P^{-1}$.

For $e^{\text{trace}(A)}$, note that $\text{trace}(A)=-3$ is just a number so $e^{\text{trace}(A)} =e^{-3}$.

Answer 3

For a $2\times2$ matrix $A$ that has distinct real eigenvalues $\lambda_1$ and $\lambda_2$ there’s a shortcut that takes advantage of the decomposition $$A=\lambda_1P_1+\lambda_2P_2 = \lambda_1{A-\lambda_2I\over\lambda_1-\lambda_2}+\lambda_2{A-\lambda_1I\over\lambda_2-\lambda_1}.$$ $P_1$ and $P_2$ are projections onto the respective eigenspaces that have the property $P_1P_2=P_2P_1=0$. Since for any projection $P$, $P^2=P$ this means that $$A^k = \lambda_1^k{A-\lambda_2I\over\lambda_1-\lambda_2}+\lambda_2^k{A-\lambda_1I\over\lambda_2-\lambda_1}$$ and $$e^{tA} = e^{\lambda_1t}{A-\lambda_2I\over\lambda_1-\lambda_2}+e^{\lambda_2t}{A-\lambda_1I\over\lambda_2-\lambda_1}$$