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For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find $$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is $$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ Notice that the assumption is implicitely used in that it implies $g>0$.

Edit. The formula above seems to be a particular case of a more general one. Suppose $A$ is $n\times n$. With Cayley-Hamilton, we have $$e^{tA}=f(t)I_n+g(t)A+\cdots+h(t)A^{n-1}.$$ Let us form the Hankel matrix $M(t)=\left(h^{(i+j-2)}(t)\right)_{1\le i,j\le n}$. Then $\det M(t)=(-1)^{n+1}e^{t{\rm Tr}\,A}$.

For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find $$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is $$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ Notice that the assumption is implicitely used in that it implies $g>0$.

Edit. The formula above seems to be a particular case of a more general one. Suppose $A$ is $n\times n$. With Cayley-Hamilton, we have $$e^{tA}=f(t)I_n+g(t)A+\cdots+h(t)A^{n-1}.$$ Let us form the Hankel matrix $M_h(t)=\left(h^{(i+j-2)}(t)\right)_{1\le i,j\le n}$. Then $\det M_h(t)=(-1)^{n+1}e^{t{\rm Tr}\,A}$.

Remark that a smooth function $h$ satisfies a linear ODE of order $n-1$ with some constant coefficients if, and only if, $\det M_h\equiv0$.

For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find $$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is $$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ Notice that the assumption is implicitely used in that it implies $g>0$.

For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find $$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is $$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ Notice that the assumption is implicitely used in that it implies $g>0$.

Edit. The formula above seems to be a particular case of a more general one. Suppose $A$ is $n\times n$. With Cayley-Hamilton, we have $$e^{tA}=f(t)I_n+g(t)A+\cdots+h(t)A^{n-1}.$$ Let us form the Hankel matrix $M(t)=\left(h^{(i+j-2)}(t)\right)_{1\le i,j\le n}$. Then $\det M(t)=(-1)^{n+1}e^{t{\rm Tr}\,A}$.

For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)I_2$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find $$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is $$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ Notice that the assumption is implicitely used in that it implies $g>0$.

For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find $$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is $$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ Notice that the assumption is implicitely used in that it implies $g>0$.