Linear operator $T$ on a finite dimensional vector space Question:

A linear operator $T$ on a finite dimensional vector space is
  diagonalizable if and only if the multiplicity of each eigenvalue
  $\lambda$ equals the dimension of $E_\lambda$.

According to me the answer should be $true$, but my book says otherwise, and so, I know that I am probably wrong, but I can't figure out, what am I missing ?
Any help will be appreciated.
 A: I think I just understood what the author(s) meant.
If you consider a linear operator $T\colon V \to V$ where $V$ is a finite-dimensional vector space over a field $k$ that is not algebraically closed, then not all zeros of the characteristic polynomial $\chi_T$ of $T$ need to lie in $k$.
If only the zeros of $\chi_T$ that lie in $k$ are considered eigenvalues, then indeed $T$ need not be diagonalisable if the dimension of $E_\lambda$ equals the multiplicity of $\lambda$ as a zero of $\chi_T$, for example consider $T \colon \mathbb{R}^3 \to \mathbb{R}^3$ given by multiplication with
$$\begin{pmatrix}0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}.$$
The characteristic polynomial is
$$\chi_T(X) = X^3 - X^2 + X - 1 = (X-1)(X^2+1) = (X-1)(X-i)(X+i),$$
and the only real eigenvalue is $1$, the dimension of the corresponding eigenspace is $1$, as is the multiplicity of the eigenvalue. But $T$ is not diagonalisable over $\mathbb{R}$ (since it has non-real eigenvalues).
If you consider every zero of the characteristic polynomial (in an algebraic closure of $k$) as an eigenvalue, then you have the simple characterisation that $T$ is diagonalisable if and only if $\dim E_\lambda$ equals the multiplicity of $X-\lambda$ in the characteristic polynomial. (And things like this are why a lot of people consider every zero of the characteristic polynomial an eigenvalue; the theory becomes much simpler then.)
