# How to deduce third eigenvalue/eigenvector pair of a matrix from first two pairs and determinant

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?

• The product of the three eigenvalues is the determinant. – user296602 Dec 11 '18 at 2:25
• I know that, but how do you find its corresponding eigenvector? – BuluBestTapu Dec 11 '18 at 2:26
• 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) – Morgan Rodgers Dec 11 '18 at 2:28
• 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. – BuluBestTapu Dec 11 '18 at 2:31
• There’s no way to determine the missing eigenspace without more information about $A$. – 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 $$(a_{i,j})$$.
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".