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Sorry if this is a dumb question but given a general 3x3 matrix

$A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} $

and assuming it has 3 distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$, is there a general (analytical) formula for the eigenvectors of this matrix?

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  • $\begingroup$ Eigenvectors are not unique to the eigenvalues. If you see the relationship between eigenvalues and eigenvectors, it is actually one to many relationship with respect to a given matrix. Hence there is no analytical formula. $\endgroup$
    – AxyuS
    Jul 17, 2020 at 17:54

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I assume the underlying field is $\mathbb{R}$.

In the case you describe - distinct eigenvalues - the vector product $$ v_{\lambda}:=(d, e-\lambda, f)\wedge (g,h,i-\lambda) $$ is a $\lambda$-eigenvector for $\lambda=\lambda_1,\lambda_2,\lambda_3$.

This is because it is perpendicular to $(d, e-\lambda, f)$, and $(g,h,i-\lambda)$; and also to $(a-\lambda, b,c)$ - which is a linear combination of $(d, e-\lambda, f)$, and $(g,h,-\lambda)$ in the case when $\lambda$ is an eigenvalue.

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  • $\begingroup$ This is so elegant! I see that you use the fact that $det(A-\lambda I)=0$ to get linear dependence of the rows of $(A-\lambda I)$. Does this argument hold if the eigenvalues are complex too? $\endgroup$
    – Sam
    Jul 17, 2020 at 22:00
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If a matrix with order $n\times n$ has $n$ distinct values then corresponding to each eigen values you will get linearly independent eigen vectors. And then you can say matrix is diagonalizable. As far as i know there is no particular formula to calculate eigen vectors for matrices. There is a general method to calculate eigen vectors that can be found in any linear algebra book.

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