# If you can multiple linearly independent eigenvectors for the same eigenvalue, why do we solve for just one?

The way I have been taught to solve for eigenvalues and eigenvectors in linear algebra is thus: compute the eigenvalues using the determinant of A-\I. Then use that to compute the eigenvectors for each eigenvalue using (A-\I)x = 0. The way I understood it, each eigenvector served as a basis for the space that contained all eigenvectors of that eigenvalue. However, I just learned from the interwebs that an eigenvalue can have more than one linearly independent eigenvector. So, what's the point of calculating just one of these for each eigenvalue, if there are theoretically infinitely many linearly independent eigenvectors for each eigenvalue? Why do we choose just one?

If you were taught that, then you were taught wrong. If $$A$$ is a $$n\times n$$ matrix then, for each eigenvalue $$\lambda$$ of $$A$$, the dimension of the space$$E_\lambda=\{v\in\mathbb{R}^n\,|\,A.v=\lambda v\}$$can go from $$1$$ to $$n$$. And, of course, if it's greater than $$1$$, a single vector $$v\in E_\lambda$$ will not be a basis of it. If the dimension is $$k$$, every basis will have $$k$$ linearly independent eigenvectors (but, unlike what you wrote, never infinitely many).
In the low-dimensional case, you start by finding the roots of the characteristic polynomial. For each root $$\lambda$$, writing $$A-\lambda I=0$$ gives you a matrix form of a linear system, and the solutions to this linear system represent the eigenspace for $$\lambda$$. The dimension of the eigenspace is at most the multiplicity of the root $$\lambda$$, which means in particular it will be finite-dimensional. And in fact there are only finitely many eigenvalues, which means there are only finitely many linearly independent eigenvectors.