I have been at this problem for a couple of hours now, this is one of the problems in the textbook which I am practicing for my midterm tomorrow.
Here's the problem:
Given a system of non-homogeneous differential equation of the form: $\vec{x}' = A\vec{x} + \vec{v}e^{\lambda t}$, where $\vec{v}$ is an eigenvector of A with eigenvalue $\lambda$. Suppose also that $A$ has $n$ linearly independent eigenvectors $\vec v_1,...,\vec v_n$ with distinct eigenvalues $\lambda_1,...,\lambda_n$ respectively.
(a) Show that this has no particular solution of the form: $\psi(t)=\vec{a}e^{\lambda t}$,
Hint: since $\{\vec v_1,....,\vec v_n\}$ is a basis for $\mathbb{R}^n$ write $\vec{a}=a_1\vec v_1+...+a_n\vec v_v$.
Here is what I have done so far:
Assume particular solution of the form: $\psi=\vec{a}e^{\lambda t}$, then: $\psi'=\vec{a}\lambda e^{\lambda t}$.
Plug these into the original equation and doing some simplification yields: $\vec{a}(A-\lambda I)=-\vec{v}$
I know that the matrix ($A-\lambda I$) does not have an inverse because $(A-\lambda I)^{-1}=\frac{adjugate(A-\lambda I)}{det(A-\lambda I)}$, since since $\lambda$ comes from: $det(A-\lambda I)=0$, therefore $(A-\lambda t)^{-1}=\frac{adjugate(A-\lambda I)}{0}$ which cannot be computed, so you cannot find $\vec{a}$, since you cannot find $\vec{a}$, you cannot guess a particular solution of the form $\psi=\vec{a}e^{\lambda t}$.
But how do I show that particular solution of the form $\psi=\vec{a}e^{\lambda t}$ cannot be used using the hint provided.
Thank you.
The equation that you almost correctly derived, $$(A−λI)\,\vec a=-\vec v,$$ along with the task construction, tells you that $$ \vec v\in \ker (A−λI)\cap \text{im}(A−λI). $$ This actually is possible for non-simple eigenvalues, for instance consider $$ \vec x{\,}'=\pmatrix{λ&1\\0&λ}\,\vec x+\vec e_1\,e^{λt} $$ then $\vec a=-\vec e_2$ gives a solution $\vec x(t)=\vec a\,e^{λt}$.
However, by the corrected assumptions all eigenvalues in your task are simple.