When solving eigenvalue equation, we usually have three scenarios. The first one is the simplest: i.e. it's diagonalised, so the eigenvectors can be found by inspection. The second one is that you know the first eigenvector and then you can find the other two by imposing orthogonal condition.
The third one is when you get matrix like
Impose det (M-lamdaI)=0 You will get eigenvalues = 1 (degenerate) and 4. My question is If I started finding eigenvector with eigenvalues 1, I got (1,0,0). And can not find three orthogonal eigenvectors.
However, if I start finding eigenvector with eigenvalue 4, I got (1,1,1). And I further imposed x = 0 , to get (0,1,-1) and finally impose orthogonality to the previous two eigenvectors to get (-2,1,1).
N.B. The above eigenvectors are not normalised.
Is it necessary to solve the problem in this order?
Geometrically, what I did above is I first found the normal of a plane, and impose x=0, to find another two orthogonal eigenvectors.
I believe for the case with 3 distinct eigenvectors, there is no preference in order when solving it.
My question is does it have to be solved in this order? Am I making it more complicated?
Orthogonality is not required.
Working with $\lambda=4$, you find a simple solution $v_0=(1,1,1)$ or any nonzero multiple.
Working with $\lambda=1$, you find a degenerate case that reduces to $x+y+z=0$. Then you can arbitrarily choose two linearly independent solutions such as $v_1=(0,1,-1)$ and $v_2=(1,0,-1)$ or any independent combinations of these.
As you see, $v_0$ is implicitly orthogonal to both $v_1$ and $v_2$, because the corresponding Eigenvalues differ.
If you want to, you can make $v_1$ and $v_2$ orthogonal, for instance by the Gram-Schmidt process, but $v_0$ needn't enter into play (orthogonality to $v_0$ is conserved).
$$v_2'=v_2-(v_1\cdot v_2)\frac{v_1}{v_1^2}=(1,-\frac12,-\frac12).$$
Now, $v_0$, $v_1$ and $v_2'$ are pairwise orthogonal. So you can really treat separately the Eigenvectors corresponding to each distinct Eigenvalue.