# Linear systems of ODEs and left eigenvalues

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).

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I am unfamiliar with what "calculate ... along ..." might mean. But this much can be said: $v^tx$ is a scalar valued function ($1 \times n$ vector times $n \times 1$ vector is a $1 \times 1$ vector). Next, $(v^tx)'=(v^t)'x+v^tx'=v^tx'=v^tAx=v^t\lambda x=\lambda v^tx$. Therefore, letting $y=v^tx$, we have $y'=\lambda y$ and so $y=Ce^{\lambda t}$. Thus $v^tx=Ce^{\lambda t}$ (for whatever that's worth). –  Bill Cook Dec 2 '11 at 19:46
@BillCook Thanks for the elucidating comment. Although one thing is still unclear to me: We does the equality $(v^t x)'=(v^t)'+v^tx'$ hold ? I thought the Leibniz rule for matrices holds only if we deal with square matrices... –  user19822 Dec 3 '11 at 11:19
The Leibniz rule holds for arbitrary matrix products. And since $(v^t)'=0$ (because $v^t$ is a constant vector), what I wrote follows. –  Bill Cook Dec 3 '11 at 14:20

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 $x$.
Thanks; I meant eigenvector of course, but I was so tired, so I mistyped it. It's corrected now. And is there nothing more to the geometric interpretation than saying that $(v^t x)$ is just a scaled exponential function ? This seems somehow shallow...is there really nothing more to it? (I have the feeling that although I understood how to get to the solution, I'm didn't get the underlying idea) –  user19822 Dec 4 '11 at 15:55
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 ... –  Robert Israel Dec 5 '11 at 1:38