4
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ It is clear ir you consider that $A^t$ is the matrix of the dual linear map. $\endgroup$
    – Leo Alonso
    Commented Jul 30, 2017 at 22:16
  • $\begingroup$ You can reduce this to the same question about the null spaces and bases for the null spaces. $\endgroup$
    – Douglas Zare
    Commented Jul 31, 2017 at 0:09

1 Answer 1

Reset to default
2
$\begingroup$

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 $\lambda_i$.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ What if there is no eigenbasis? $\endgroup$
    – Ted Shifrin
    Commented Jul 30, 2017 at 22:54
  • $\begingroup$ You can run basically the same argument with the expression $(A - I)^k$ and generalized eigenvectors. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 30, 2017 at 23:02
  • $\begingroup$ Since you need the eigenvalue in there, I don't quite see it quickly, @Qiaochu. $\endgroup$
    – Ted Shifrin
    Commented Jul 30, 2017 at 23:16
  • $\begingroup$ Note that, in a generalization of this argument to a non-diagonalizable matrix, if $v_1$ is an eigenvector and $v_2$ is the second vector of a Jordan basis, then their duals swap their roles: $f_2$ is the eigenvector and $f_1$ is the generalized eigenvector. I think that the generalized proof will have to account for this property in some way. $\endgroup$
    – Federico Poloni
    Commented Aug 30, 2017 at 8:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .