Sorry for the long prose.

I am trying to understand a naive treatment of the solution to the Linear System with constant coefficients $$ \left({ \begin{matrix} y_1(t) \\ y_2(t) \\ y_3(t) \\ \end{matrix}} \right) = A_{3 \times 3} \left({ \begin{matrix} y_1'(t) \\ y_2'(t) \\ y_3'(t) \\ \end{matrix}} \right) $$

when the eigenvalues of $A$ may be defective.

Case 1: $A$ has only one eigenvalue $\lambda$ of multiplicity $ 3$ but with geometric multiplicity $1$ - that is space of corresponding eigenvectors of $\lambda$ is of dimension $1$. My notes tell me we need to look for solutions of the form $$Y_1 = X_1 e^{\lambda t} $$ $$ Y_2 = [X_1 t + X_2] e^{\lambda t} $$ $$ Y_3 = [X_1 \dfrac{t^2}{2} + X_2 t + X_3 ] e^{\lambda t} $$

where the equations $ (A - \lambda I)X_3 = X_2, \;\; (A - \lambda I)X_2 = X_1, \;\; (A - \lambda I)X_1 = 0 $ hold for non-zero $3 \times 1$ constant vectors $X_i$. I understand how the solutions are justified. But,

How to prove that $Y_1, Y_2, Y_3$ are linearly independent solutions? Or is it easier to first prove $X_1, X_2, X_3$ are linearly independent?

My notes exclude this bit and I am having trouble proving these. (Haven't learnt canonical forms. Knowledge on Linear Algebra only extends to basic details on Eigenvalues i.e. how they represent Invariant Subspaces).

Then I need to extend this to the case when $A$ has two eigenvalues $\lambda_1$ and $\lambda_2$ with $\lambda_1$ having Geometric and Algebraic Multiplicity $1$ and $\lambda_2$ having algebraic multiplicity $2$ and Geometric multiplicity $1$ - (one missing solution).

Any help is appreciated.


1 Answer 1


Let $a_1\dots a_3$ such that $\sum_{i=1}^3 a_iX_i=0$.

Let $I$ be the largest index with $a_i\ne0$, i.e. $$ I = \max\{i : a_i \ne 0\}. $$ Applying $(A-\lambda I)^{i-1}$ tho the linear combination above yields $$ \sum_{i=1}^I a_i(A-\lambda I)^{i-1}X_i = a_I X_1, $$ implying $a_I=0$. A contradiction.


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