Diagonal matrices (if entries are not repeating) have eigenvectors only from the standard basis (i.e. components are: single $1$ and the rest $0$ - of course up to the scale) - I believe that there are no other matrices than diagonal ones with distinct entries which have full set of standard basis vectors (SBV) as eigenvectors.

Indeed, $D*I=D$, where identity matrix presents the full set of standard basis vectors. In the case of repeating diagonal entries also linear combinations of eigenvectors can be eigenvectors what we would like to exclude from the set of eigenvectors as a condition imposed on the problem.

The same can be said about Jordan forms of matrices, they also have eigenvectors from the standard basis, but the set has now a less number of vectors, it seems that for every Jordan block we can have only one eigenvector (distinctivness of eigenvalues between different blocks is also important here, I suppose).

If we write for Jordan block $J* I=J$ it is visible that it has only one possible eigenvector from the standard basis. The same should be for powers $J^k$ of Jordan blocks.

  • I'm interested in the problem what is the general form of a matrix that has eigenvectors only from the standard basis. Are there others besides diagonal and Jordan forms?

Knowing such forms would allow to present matrices with different geometric and algebraic multiplicities with easy in form eigenvectors. I suppose that necessary condition for such matrix is its upper(lower) trianguality.

The difficulty of problem, it seems, doesn't lie here in proving that for a supposed matrix some vector from the standard basis belongs to the set of eigenvectors, but that there are no other eigenvectors, that are different from SBV.

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    $\begingroup$ Of course not. Take a Jordan matrix, $J$, and its eigenvalues. Then take any isomorphism $P$ that sends those eigenvectors to elements of the standard basis. The matrix $P^{-1}JP$ has the same property. These are all the matrices with that property. $\endgroup$ – orole Jan 17 '18 at 15:55
  • $\begingroup$ The identity matrix is diagonal and has eigenvectors which aren't basis vectors ... $\endgroup$ – Arthur Jan 17 '18 at 15:55
  • $\begingroup$ @Arthur Of course the identity matrix is specific $\endgroup$ – Widawensen Jan 17 '18 at 15:58
  • $\begingroup$ @orole Could you give an example? $\endgroup$ – Widawensen Jan 17 '18 at 16:00
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    $\begingroup$ Well, we also have $\left(\begin{smallmatrix}2&0&0\\0&2&0\\0&0&5\end{smallmatrix}\right)$. This time $(1,1,0)$ is an eigenvector with eigenvalue $2$, but not a basis vector. $\endgroup$ – Arthur Jan 17 '18 at 16:00

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