# question on left and right eigenvectors

I know that $\text{A}\textbf{x}=\lambda \textbf{x}$, where $\textbf{x}$ is right eigenvector, while in $\textbf{y}\text{A}=\lambda \textbf{y}$, $\textbf{y}$ is left eigenvector.But What is significance of left and right eigenvectors ? How do they differ from each other geometrically?

-

The set of left eigenvectors and right eigenvectors together form what is known as a Dual Basis and Basis pair.

http://en.wikipedia.org/wiki/Dual_basis

In simpler terms, if you arrange the right eigenvectors as columns of a matrix B, and arrange the left eigenvectors as rows of a matrix C, then BC = I, in other words B is the inverse of C

-

Geometrically the matrix $A$ is an origin and line preserving transformation (${\bf v}\mapsto A\cdot{\bf v}$). The right eigenvectors are eigenvectors for this transformation, but the left ones for $A^T$, which, geometrically can be totally different.

However, the eigenvalues and the dimensions of their corresponding eigenspaces must stay the same.

-
"Origo" looks like something that's having difficulty coming across in translation... can you explain what that means? "Origin" maybe? –  rschwieb Sep 21 '12 at 16:44

Using $A$ as a linear transformation on the right or on the left produces (in general) two completely different transformations of the vector space.

These two transformations have their own eigenvectors, which may have nothing to do with each other.

The geometric significance of eigenvectors is: they lie in subspaces which are stretched by $A$, but not tilted at all.

-

The (right) eigenvectors for $A$ correspond to lines through the origin that are sent to themselves (or $\{0\}$) under the action $x\mapsto Ax$. The action $y\mapsto yA$ for row vectors corresponds to an action of $A$ on hyperplanes: each row vector $y$ defines a hyperplane $H$ given by $H=\{\text{column vectors }x: yx=0\}$. The action $y\mapsto yA$ sends the hyperplane $H$ defined by $y$ to a hyperplane $H'$ given by $H'=\{x: Ax\in H\}$. (This is because $(yA)x=0$ iff $y(Ax)=0$.) A left eigenvector for $A$, then, corresponds to a hyperplane fixed by this action.

-