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

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The discussion focuses on proving that a set of functions forms a basis for the solution space of a homogeneous linear differential equation with constant complex coefficients. The solution space is associated with a polynomial whose roots are distinct complex numbers. The speaker is using induction to demonstrate the linear independence of the proposed basis set, encountering difficulty in the induction step. A hint suggests applying the operator (D - c_mI)^{n_m} to a linear combination of the basis functions to show that it results in zero, indicating that all coefficients must be zero. The key point is that this operator maps functions in the subspace spanned by the basis to zero, supporting the proof of linear independence.
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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.
 
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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.
 
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.
 

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