Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?

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    $\begingroup$ The product of the three eigenvalues is the determinant. $\endgroup$ – user296602 Dec 11 '18 at 2:25
  • $\begingroup$ I know that, but how do you find its corresponding eigenvector? $\endgroup$ – BuluBestTapu Dec 11 '18 at 2:26
  • $\begingroup$ Same as you find any eigenvector: Null space of $A-\lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors) $\endgroup$ – Morgan Rodgers Dec 11 '18 at 2:28
  • $\begingroup$ You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this. $\endgroup$ – BuluBestTapu Dec 11 '18 at 2:31
  • $\begingroup$ There’s no way to determine the missing eigenspace without more information about $A$. $\endgroup$ – amd Dec 11 '18 at 2:50

Let $A=[a_{i,j}]$. You know $spectrum(A)=\{\lambda,\mu,\nu\}$ and $u,v\in\mathbb{R}^3$ s.t. $Au=\lambda u,Av=\mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.

So, you miss 2 relations.

EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".

Indeed, the eigenvector of $A^T$ associated to $\nu$ is orthogonal to the plane $span(u,v)$.

Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".


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