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Everywhere I can read that a matrix and its adjoint have pretty much the same eigenvalues ( only complex conjugation is the difference between them). Now I was wondering whether such a relation also exists between the eigenvectors of both matrices. Do they have something in common?

glS
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    If the matrix is normal (ie commutes with its adjoint), a lot can be said, an is covered in many linear algebra texts. Otherwise, I think the relationship is less clear. – Geoff Robinson Jun 02 '13 at 07:59
  • yes, you are right. I am only interested in this general case. –  Jun 02 '13 at 08:00
  • In general, the fact that two matrices have the same eigenvalues does not necessarily imply that they have the same eigenvectors. Therefore, we cannot conclude anything from the fact that they have similar eigenvalues yet. – dreamer Jun 02 '13 at 09:03

4 Answers4

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There is a very important orthogonality relation between eigenvectors of a matrix and its adjoint which is very useful in many applications (reduction of differential equations for example). Let $A$ be a matrix and $A^{\ast}$ be its adjoint.

If $A x = \lambda x$ and $A^{\ast} y = \mu y$ with $\lambda \neq \overline{\mu}$ (where $\overline{\cdot}$ is complex conjugate) then $x$ and $y$ are orthogonal. i.e. $(x, y)= 0$.

The proof is very easy. $$ \lambda (x, y) = (A x, y) = (x, A^{\ast} y) = (x, \mu y) = \overline{\mu} (x, y), $$ which implies that $(\lambda - \overline{\mu})(x, y) = 0$. This implies the orthogonality of $x$ and $y$.

pitonist
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You are essentially asking about the relation between right and left eigenvectors of the same matrix. Basically, even though they occur for the same eigenvalues, there is not much relation between them. While right eigenvectors live in the vector space you matrix acts upon, the left eigenvectors live in the dual space (the are linear forms on the space). Even if one assumes for simplicity that the matrix is diagonalisable, then the left eigenvectors for$~\lambda$ are those linear forms that vanish on all the other eigenspaces than that for$~\lambda$; clearly knowing just the eigenspace for$~\lambda$ is not sufficient to known what that left-eigenspace is. The only positive thing I can think of is that for every left eigenvector there is some right-eigenvector for the same eigenvalue on which it does not vanish (obviously that vector is not unique).

Marc van Leeuwen
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The eigenvectors of $A^T$ are the columns of $P^{-T}$, where $P$ is the eigenvectors of $A$ (assumed diagonalisable).

This can be shown by transposing the eigendecomposition of $A = P^{-1} D P$ which is unique (up to scalings on the columns).

As a corollary, the eigenvectors of $A$ and $A^T$ of different eigenvalues are orthogonal. This follows from $(P^{-T})^T P S = P^{-1} P S = S$, where S is a diagonal matrix containing the arbitrary scalings.

Patrick
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Regarding the left and right eigenvectors, assuming that the matrix is diagonalizable: Suppose we have one eigenvalue lambda of A (NXN) and all of its eigenvectors. Find the (N-1) dimensional subspace spanned by all the eigenvectors except for the one associated with lambda. Now consider the vector perpendicular to this subspace. It will be the eigenvector of A-transpose associated with lambda. (This shows that it can be done. Not that this is an efficient way to do it computationally.)

John Finn
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