2
$\begingroup$

In a homework assignment on ODEs I'm supposed to "calculate $(v^t x)$ along x'=Ax and to interpret the result geometrically", where $v$ is the left eigenvector of $A$, meaning $v^t A= \lambda v^t$.

My question is: What does is mean to calculate $(v^t x)$ "along" something ? We did some theory on systems of linear equations, but we have never calculated "something along something". Could you explain to me how to attack this problem ?

I'm also puzzled by the use of left eigenvectors (In none of the classes I took so far this concept was mentioned): Why does one even bother to define it like this ? An eigenvector is something that belongs to an operator so it seems kind of unnatural for me, to define it as left/right for matrices, for two reasons:

1) for operators there exist only a "right" eigenvectors, $\mathcal{A}(v)=\lambda u $.

2) $v^t A= (A^t v)^t$, so one can always reduce a left eigenvectors to a right one, by transposing the matrix, so defining left eigenvectors seems somehow pointless (please correct me if I'm wrong).

  • 0
    The Leibniz rule holds for arbitrary matrix products. And since $(v^t)'=0$ (because $v^t$ is a constant vector), what I wrote follows.2011-12-03

1 Answers 1

1

Actually $v$ is a left eigen*vector*, not eigen*value*.

It simply means calculate $v^t x$ where $x$ is a solution of x' = A x.

Hint: multiply $v^t$ by both sides of the differential equation and see what you get.

The reason you're using left eigenvectors is that there's already something on the right of $A$, namely $x$.

  • 0
    As a geometric interpretation, you might say that $x \to v^t x$ is the component of $x$ in a certain direction; $v$ being a left eigenvector means that $A$ multiplies this component of any vector by $\lambda$; and so the diferential equation says that this component of the vector will grow exponentially ...2011-12-05