My book writes the following statement:
For a square matrix $A$ of order $n$ to be diagonalizable, the sum of the dimensions of the eigenspaces must be equal to $n$. One way this can happen is when $A$ has $n$ distinct eigenvalues.
I'm a little confused on the last sentence. Are they trying to say that if $A$ has $n$ eigenvalues, then it must be the case that each eigenspace is $1$-dimensional? Couldn't it be possible for one or more of the eigenspaces to be $2$-dimensional or three-dimensional, etc?
Best Answer
The intersection of the eigenspaces corresponding to different eigenvalues is always $\{0\}$. Indeed if $x \in E_m$, the eigenspace of $m$ and $x \in E_n$ then we have $Ax = mx$ and $Ax = nx$ so $(m-n)x = 0$ so, since $m \neq n$, $x$ must be $0$. Now let $d_i$ be the dimension of the eigenspace of eigenvalue $m_i$ with $i \in \{1 \ldots n\}$ then the only way $d_1+d_2 + \ldots +d_n = n$ is when all the $d_i = 1$.