# General formula for eigenvectors of a 3x3 matrix

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?

• 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. Jul 17, 2020 at 17:54

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.
• 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?
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.