Toronto Math Forum
MAT2442018F => MAT244Lectures & Home Assignments => Topic started by: Karim El Komaishy on November 07, 2018, 11:19:35 PM

When I tried some of the questions that involved matrices with complex eigenvalues, I sometimes ended up with matrices whose reduced row echelon forms that have pivot positions on each column (after subracting the main diagonal entries with the eigenvalues; no free variables). Does this sometimes happen whenever I try to find eigenvectors with this?

Not c lear what do you mean.May be an example?

Not really I think. We have to get a matrix contains nonzero number on the first row and zero on the second row in reduced row echelon forms.

There should always be at least one free variable if eigenvalue is chosen correctly. Free variable need to be present so that your solution is a line or plane or etc.

If the characteristic polynomial $\det (AkI)$ has a multiple root (say root $k$ of multiplicity $m$) then it can have any number $l=1,\ldots, m$ of linearly independent eigenvectors, associated with this eigenvalue (we do not distinguish here real or complex eigenvalues). Then one needs to find Jordan (echelon?) form.
In particular, let $m=2$. Then either there are two l.i. eigenvectors $\mathbf{e}_1$, $\mathbf{e}_2$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \mathbf{e}_1+C_2\mathbf{e}_2\Bigr)
$$
or there exists a vector $\mathbf{f}$ s.t. $(Ak)^2\mathbf{f}=0$, but $\mathbf{e}=(Ak)\mathbf{f}\ne 0$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}+\mathbf{f}\bigr]+C_2\mathbf{e}\Bigr)
$$
If $m=3$ then there are three cases: either there are three l.i. $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$, in which case solution is similar to the former case of $m=2$, or there are a vector $\mathbf{f}$ s.t. $(Ak)^2\mathbf{f}=0$, but $\mathbf{e}_1=(Ak)\mathbf{f}\ne 0$, and $\mathbf{e}_2$, also eigenvector but not proportional $\mathbf{e}_1$; then the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[t\mathbf{e}_1+\mathbf{f}\bigr]+C_2\mathbf{e}_1+C_3\mathbf{e}_2\Bigr);
$$
or there exists a vector $\mathbf{g}$ s.t. $(Ak)^3\mathbf{g}=0$, but $\mathbf{e} =(Ak)^2\mathbf{g}\ne 0$; let $\mathbf{f} =(Ak)\mathbf{g}$ and
the corresponding part of the solution is
$$
e^{kt}\Bigl( C_1 \bigl[\frac{t^2}{2}\mathbf{e}_1+t\mathbf{f}+\mathbf{h}\bigr]+C_2\bigl[t\mathbf{e} +\mathbf{f}\bigr]+C_3\mathbf{e}\Bigr).
$$
and so on...