I have to show that the dynamic matrix of a time-invariant discrete-time system has eigenvalues located at $1+\lambda_cT$, if the discretization was performed by the explicit Euler method. This should be shown for an arbitrary $n \times n$ matrix $A$. For the explicit Euler method the dynamic matrix is computed via $I + AT$, where:
$I$ ...the identity matrix
$\lambda_c$ ...eigenvalues of the continuous time system
$A$ ...dynamic matrix
$T$ ...discretization time.
My approach:
For the eigenvalues, we want to compute the determinants and set them to zero:
$$ det((I + AT) - \lambda_d I) = 0 $$
for the discrete system, where $\lambda_d$ denotes the eigenvalues of the discrete case. Then we have:
$$ det(A - \lambda_cI) = 0 $$
for the continuous system.
I thought of just comparing the two equations and solve for $\lambda_d$, to get $1 + \lambda_cT$, which does not quite work. I am bit stuck here and would be glad about any hint.
Substituting $\lambda_d=1+\lambda_c\,T$ into expression inside the determinant of the equation for the discrete time eigenvalue problem yields
\begin{align} (I + A\,T) - \lambda_d\,I &= (I + A\,T) - (1+\lambda_c\,T)\,I \\ &= I + A\,T - I - \lambda_c\,T\,I \\ &= (A - \lambda_c\,I)\,T. \end{align}
So as long as $T\neq0$ then $\det((I + A\,T) - \lambda_d\,I)=0$ is equivalent to $\det(A - \lambda_c\,I)=0$ when using $\lambda_d=1+\lambda_c\,T$.