If an $n \times n$ matrix has $n$ distinct eigenvalues, does each of its eigenspaces have to be $1$-dimensional? 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?
 A: 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 $d_i = 1$.
A: Well if it has n distinct eigenvalues then yes, each eigenspace must have dimension one. This is because each one has at least dimension one, there is n of them and sum of dimensions is n, if your matrix is of order n it means that the linear transformation it determines goes from and to vector spaces of dimension n.
If you have 2 equal eigenvalues then no, you may have a eigenspace with dimension greater than one.
A: Remember that, dim(U+V) = dim(U) + dim(V) - dim(U ∩ V)
If a matrix A has 'n' distinct Eigenvalues, then the matrix A has 'n' Linearly Independent Eigen Vectors.
Let the eigenspace formed by λi is Ei.
A vector x can not belong to two different Eigenspaces. And thus making the intersection of the eigenspaces trivial. Thus, summing the eigenspaces corresponding to 'n' distinct eigenspaces:
dim(E1+E2+E3+...+En)=dim(E1)+dim(E2)+dim(E3)+...+dim(En)
=> n = dim(E1)+dim(E2)+...+dim(En)
And since dim(Ei)>0, it follows that dim(Ei) is indeed equal to 1.
