# Conceptual understanding of eigenvalues and eigenvectors

I am just getting started with Linear Algebra and I have a conceptual doubt regarding eigenvalues and eigenvectors.

Suppose we have $$Ax = \lambda x$$ where $$\lambda$$ is a scalar , $$x\not= 0$$ is called the eigenvector of matrix $$A$$ corresponding to eigenvalue $$\lambda$$. If we multiply $$x$$ by some scalar $$\alpha$$ times then we have that

$$A(\alpha x)=\alpha Ax = \alpha \lambda x = \lambda (\alpha x)$$

Now we have that $$(\alpha x,\lambda)$$ is an eigenvector-eigenvalue pair of $$A$$. Then isn't it the case that we have found all the eigenvectors just by taking scalar multiples of eigenvectors? In other words, why do we specifically solve the equation $$det(A-\lambda I)=0$$ if we can get them as above?

• Some eigenspaces are 2-dimensional or higher. Scaling alone won't find all the eigenvectors. Oct 5, 2021 at 12:58

## 2 Answers

We solve the equation $$\det(A-\lambda\operatorname{Id})=0$$ in order to know which scalars turn out to be eigenvalues. Then, after having them, for each each $$\lambda$$ which turns out to be an eigenvalue, we search for the corresponding eigenvectors by solving the equation $$Av=\lambda v$$.

What you have found is that if $$x$$ is an eigenvector of the eigenvalue $$\lambda$$ than also $$\alpha x$$ is an eigenvector of the same eigenvalue. In other words: to a given eigenvalue corresponds an entire linear space of eigenvectors, spanned by $$x$$. This is called the eigenspace of $$\lambda$$. But, for a different eigenvalue the eigenspace is, in general, different, and can be determined only if you know the eigenvalue.