Concerning the general solutions to linear ODEs Y'=AY when A has multiple eigenvalues

Question

Given linear ODES Y'=AY, where Y is a column vector, A is a 6*6 square matrix. Clearly A has 6 eigenvalues, namely r1, r2, r3, r4, r5, r6. Herein we assume r5=r2, r6=r3.That is, r2 and r3 are two couple eigenvalues. The problem is how to obtain the universal solutions (general solutions) to Y'=AY. As usual, we can use mathematica or matlab to obtain r1, r2, r3, r4 and their corresponding four independent eigenvectors v1, v2, v3 and v4. The next step is how to get the other two eigenvectors relating to r5 (r2) and r6 (r3). Even if we got the other two vectors say v5, v6. Then how to present the general solutions? Can it be expressed as {v1*exp[r1*x],v2*exp[r2*x],v3*exp[r3*x],v4*exp[r4*x],v5*exp[r2*x],v6*exp[r3*x]}*C, where C={c1, c2, c3, c4, c5,c6}.c1,..., c6 are six undetermined constants. Otherwise can we use MatrixExp for Ax (i.e., fundamental solution matrix) to express the general solutions? Is it in the form of MatrixExp[Ax]*C? You know, this problem has bothered me for over two years. Thanks so much.

Answer 1

So you have $4$ eigenvalues where two of them have multiplicity $2$, namely $r_2$ and $r_3$. There are two cases to consider.

If the eigenspace corresponding to $r_2$ and $r_3$ each have dimension $2$, then you can obtain $2$ linearly independent eigenvectors corresponding to each of $r_2$ and $r_3$. Thus there is a basis of eigenvectors for the whole $6$ dimensional space and so you can diagonalize the matrix. Say $A = P^{-1}DP$ for an invertible matrix $P$ and diagonal matrix $D$. Then the solution is $Y(t) = e^{At}c = e^{P^{-1}DPt}c = P^{-1}e^D Pc$ (where $c$ some constant vector) and calculating $e^D$ is trivial. Note that the last equality can be obtained by using the series definition of the exponential of a matrix.

On the otherhand, if at least one of the eigenspace corresponding to $r_2$ and $r_3$ does not have dimension $2$, then you cannot form a basis of eigenvectors for the whole space. So the matrix $A$ is not diagonalizable. However, you can always find its Jordon canonical form (with possibly complex entries). Then you can proceed as before but calculating $e^D$ is more involved where $D$ is the Jordon canonical form for $A$.

Answer 2

On the other hand, actually I can get another two independent vectors say v5 and v6 using the knowledge of textbooks for ODES. However, I don't know how to express the general solution by using the obtained eigenvalues and corresponding eigenvectors. I am confused by the form of the general solution, Should it be {v1*exp[r1*x],v2*exp[r2*x],v3*exp[r3*x],v4*exp[r4*x],v5*exp[r2*x],v6*exp[r3*x]}*C or {{v1*exp[r1*x],x*v2*exp[r2*x],x*v3*exp[r3*x],v4*exp[r4*x],v5*exp[r2*x],v6*exp[r3*x]}*C. Note that the above two forms are slightly different cuz v2*exp[r2*x],v3*exp[r3*x] in the former form are replaced by x*v2*exp[r2*x],x*v3*exp[r3*x] in the latter one.