# Does the left and right eigenvector of a matrix corresponding to different eigenvalues ALWAYS have a zero dot product?

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.

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$\LaTeX$ 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

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 $0=\omega^tv=\left<\omega,v\right>$.

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Thanks. This was very helpful. –  Zizo Dec 11 '11 at 11:10

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 $y^T x=0$.

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