If a matrix $A$ is diagonalizable, what does this imply involving the dimensions of the eigenspaces of $A$?
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What do you think? – Aniruddha Deshmukh Nov 05 '21 at 13:55

@AniruddhaDeshmukh , I know that when A is diagonalizable, then it must have n distinct eigenvalues that correspond to linearly independent eigenvectors. I do not know what this implies regarding the eigenspaces of $A$. – Kaixzer Nov 05 '21 at 14:04

What is the definition of eigenvalues and eigenvectors and eigenspaces? – Aniruddha Deshmukh Nov 05 '21 at 14:05

I am having a hard time understanding eigenspaces. The definition given to me is that the eigenspace is the null space of $(A \lambda I)$ which I find as somewhat unclear. – Kaixzer Nov 05 '21 at 14:21

Simply put, eigenspace corresponding to an eigenvalue $\lambda$ is the collection of all vectors $x$ such that $Ax = \lambda x$. Clearly, if $x$ is an eigenvector with eigenvalue $\lambda$ then it is in the eigenspace of $\lambda$ and conversely, any vector in the eigenspace is either an eigenvector or the zero vector. With this understanding, if you fix an eigenvalue, what can you say about the dimension of its eigenspace? – Aniruddha Deshmukh Nov 05 '21 at 14:24

Thanks. I've got it now. – Kaixzer Nov 05 '21 at 14:46