Theorem 6.6: Let $A$ be a square matrix, let $\gamma$ be an eigenvalue of $A$ with multiplicity $m$. Then the dimension of the eigenspace of $A$ corresponding to the eigenvalue $\gamma$ is less than or equal to $m$.
Could someone explain the intuition behind this theorem? I am not looking for a proof, one already exists on this site. Please give me intuition behind why its true.
Related: Eigenspace and polynomials?
Thanks.
Best Answer
Here is how I think about this: consider the case of a nilpotent operator $A$ on a vector space $V$ of dimension $n$, so that $A^n=0$. The characteristic polynomial is $T^n$. The $0$-eigenspace of $A$ is just its kernel; thus, in this case, the theorem says that the dimension of the kernel of $A$ is not more than the dimension of $V$, which is obvious. In fact, the essential argument is already contained in this simple case.
You can view it like this: the evil operator $A$ is trying to kill the innocent citizens of $V$, but they're not going without putting up a fight. The unlucky citizens die at time $t=1$: those are the ones in the kernel of $A$. The luckiest citizens might survive up to time $t=n$. The theorem says that the set of unlucky citizens is contained in the set of all citizens.