Proving a basis is a basis for homogeneous linear differential equation with constant (complex) coefficients

[psie]

TL;DR

I'm working an exercise to prove a certain theorem in my linear algebra book. I'm stuck with a computation of verifying linear independence.

I struggle with a certain computation. Consider a homogeneous linear differential equation with constant (complex) coefficients. Such an equation is associated with a polynomial $p(t)$, which we can write $$p(t)=(t-c_1)^{n_1} (t-c_2)^{n_2}\cdots (t-c_k)^{n_k},$$where $n_1,n_2,\ldots,n_k$ are positive integers and $c_1,c_2,\ldots,c_k$ are distinct complex numbers. A basis for the solution space is $$S=\{e^{c_1t},te^{c_1t},\ldots,t^{n_1-1}e^{c_1t},\ldots,e^{c_kt},te^{c_kt},\ldots,t^{n_k-1}e^{c_kt}\}.$$ I'm in the process of proving that $S$ is indeed a basis for the null space of $p(\mathsf D)$, where $\mathsf D$ is the differential operator. I want to proceed by induction on $k$ to prove linear independence. I can prove the base case but am stuck on the induction step. Suppose $S$ is linearly independent for all $k<m$ and suppose $$\sum_{i=1}^m\sum_{j=0}^{n_i-1}b_{ij}t^je^{c_it}=0.\tag1$$We want to show $b_{ij}=0$ for all $i,j$. Now I've received a hint to apply $(\mathsf D-c_m\mathsf I)^{n_m}$ to $(1)$, i.e. $$(\mathsf D-c_m\mathsf I)^{n_m}\left( \sum_{i=1}^m\sum_{j=0}^{n_i-1}b_{ij}t^je^{c_it}\right)=0.$$Any tips and tricks on how to proceed? I struggle with how to simplify the above.

 

[fresh_42]

This might be a silly question, but what you have is $\sum n_i$ many terms all of the form $t^ke^{ct}.$ Every constellation $b_1t^ke^{ct}+b_2t^me^{dt}=0$ yields $b_1=b_2=0$ by substituting $t\in \{0,1\}$ except for the case $c=d$ and $k=m.$

The induction step is accordingly.

 

[pasmith]

The point here is that (D - c_mI)^{n_m} is the zero map on the subspace E_m spanned by \{ t^ke^{c_mt} : k = 0, 1 , \dots, n_m - 1 \}. This is so because <br /> (D - c_mI)(t^ke^{c_mt}) = kt^{k-1}e^{c_mt} and thus D - c_mI maps everything in E_m to zero in at most n_m iterations.