Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
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

1 Answer 1

up vote 1 down vote accepted

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 $x$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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