0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.