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.

Let $\mathcal{E} = \lbrace v^1 ,v^2, \dotsm, v^m \rbrace$ be the set of right eigenvectors of $P$ and let $\mathcal{E^*} = \lbrace \omega^1 ,\omega^2, \dotsm, \omega^m \rbrace$ be the set of left eigenvectors of $P.$ Given any two vectors $v \in \mathcal{E}$ and $ \omega \in \mathcal{E^*}$ which correspond to the eigenvalues $\lambda_1$ and $\lambda_2$ respectively. If $\lambda_1 \neq \lambda_2$ then $\langle v, \omega^\tau\rangle = 0.$

Proof. For any eigenvector $v\in \mathcal{E}$ and $ \omega \in \mathcal{E^*}$ which correspond to the eigenvalues $\lambda_1$ and $\lambda_2$ where $\lambda_1 \neq \lambda_2$ we have, \begin{equation*} \begin{split}\langle\omega,v\rangle = \frac{1}{\lambda_2}\langle \lambda_2 \omega, v\rangle = \frac{1}{\lambda_2} \langle P^ \tau \omega ,v\rangle = \frac{1}{\lambda_2}\langle\omega,P v\rangle = \frac{1}{\lambda_2} \langle \omega,\lambda_1v\rangle = \frac{\lambda_1}{\lambda_2}\langle \omega,v\rangle .\end{split} \end{equation*} This implies $(\frac{\lambda_1}{\lambda_2} - 1)\langle\omega,v\rangle = 0.$ If $\lambda_1 \neq \lambda_2$ then $\langle \omega,v\rangle = 0.$

My question: what if $ \lambda_2 = 0 \neq \lambda_1,$ how can I include this case in my proof.

share|improve this question
$\LaTeX$ tip: don't use < and > for left and right angle brackets; $\LaTeX$ treats them like operators rather than delimiters, so the spacing is all wrong. Use \langle and \rangle instead. –  Arturo Magidin Dec 10 '11 at 21:55
Thanks Arturo for your suggestion. It was very useful. –  Zizo Dec 11 '11 at 9:48
add comment

2 Answers

up vote 2 down vote accepted

Do not be afraid to use the formulation of the dot/inner product as a transpose/conjugate vector multiplied by actual vector. Recall that $v$ is a right eigenvector if and only if $Pv=\lambda_1 v$ and that $\omega$ is a left eigenvector if and only if $\omega^t P=\lambda_2\omega^t$.

Then $\left<\omega,v\right>=\omega^tv$. Sticking a $P$ in the middle and using associativity of matrix multiplication and commutativity of scalar multiplication we get $$\lambda_2(\omega^tv)=(\lambda_2\omega^t)v=(\omega^tP)v=\omega^t(Pv)=\omega^t(\lambda_1v)=\lambda_1(\omega^tv)$$

Subtracting the far sides from each other we obtain $(\lambda_1-\lambda_2)(\omega^tv)=0$, and since the scalars $\lambda_1-\lambda_2$ and $\left<\omega,v\right>=\omega^tv$ belong to a field (which in particular has no zero-divisors other than $0$ itself) we must have either $\lambda_1=\lambda_2$ or $0=\omega^tv=\left<\omega,v\right>$.

share|improve this answer
Thanks. This was very helpful. –  Zizo Dec 11 '11 at 11:10
add comment

Let $Ax=\lambda x$ and $y^T A = \mu y^T$ with $\lambda \neq \mu$. Multiply $Ax=\lambda x$ from the left by $y^T$ and you get $\mu y^T x = \lambda y^T x$ or equivalently $(\lambda - \mu) y^T x =0$. Since $\lambda - \mu \neq 0$ this yields $y^T x=0$.

Note that i assumed that we can "cancel" $(\lambda - \mu)$. This is true as long as the set of our scalars forms an integral domain. For the purposes of standard linear algebra this is always true, since the scalars form a field, and a field is an integral domain.

If say $\lambda=0$, then the relation $(\lambda - \mu) y^T x =0$ becomes $\mu y^T x =0$. Since $\mu \neq 0$ (we assume $\lambda \neq \mu$), this yields $y^T x=0$.

share|improve this answer
add comment

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.