I know that $A$ and $A^t$ have the same characteristic polynomial. But I'm looking for some picture of why they should have the same set of eigenvalues. 


Maybe slightly more concrete question is whether or not you can say something about bases of $A^t$ given an eigenbasis of $A$. Does knowing one give you a computational advantage in computing the other? On one hand, they seem unrelated. On the other hand, if you know the standard basis is an eigenbasis of $A$, then the standard basis is forced to be eigen w.r.t. $A^t$. 


Insights or references for either of these questions would be appreciated. 



geometric intuition for A and A-transpose having the same eigenvalues
Suppose $v\_j$ is an eigenbasis of $A$ with eigenvalues $\lambda\_j$, so that $A v\_j = \lambda\_j v\_j$. Then for all dual vectors $f$ we have


$$\langle f, A v\_j \rangle = \lambda\_j \langle f, v\_j \rangle$$


where $\langle -, - \rangle$ denotes the dual pairing. If $f\_i$ denotes the dual basis to $v\_j$, so that $\langle f\_i, v\_j \rangle = \delta\_{ij}$, then we get


$$\langle f\_i, A v\_j \rangle = \lambda\_j \langle f\_i, v\_j \rangle = \lambda\_j \delta\_{ij}.$$


Rewriting this as


$$\langle A^{\ast} f\_i, v\_j \rangle = \lambda\_j \delta\_{ij}$$


and fixing $i$ and varying over $j$, then varying over $i$, shows that $A^{\ast} f\_i = \lambda\_i f\_i$, hence $f\_i$ is an eigenbasis of $A^{\ast}$ with eigenvalues $\lambda\_i$. 


