Doesn't all matrices have an diagonal matrix with respect to some basis? I am reading Linear algebra done right. I was going to a Proposition 5.21. Which goes like this.
Suppose $T \in L(V)$. Let  $\lambda_1 , . . . , \lambda_n$ denote the
distinct eigenvalues of $T$ . Then the following are equivalent:
(a) $T$ has a diagonal matrix with respect to some basis of $V$ ;
(b) $V$ has a basis consisting of eigenvectors of $T$ ;
My question is, Isn't all linear operators $V \rightarrow V$ is just a change of basis? In which case, isn't there always a diagonal matrix (Identity) with the right basis? Or am I wrong somewhere? What does this some basis really mean in
Suppose $T \in L(V)$ has an upper-triangular matrix
with respect to some basis of $V$ . Then the eigenvalues of $T$ consist
precisely of the entries on the diagonal of that upper-triangular matrix. (Proposition 5.18)
 A: Consider $n = 2$ and $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. 
Obviously the eigenvalues of $A$ cannot be different from $0$. 
If $A$ were diagonizable, then there would exist an invertible matrix $V$ such that $V^{-1} A V = D$ where $D$ is a diagonal matrix containing the eigenvalues of $A$. But this implies $D = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} $, leading to the contradiction $A = D$. Thus there is no basis such that $A$ is a diagonal matrix (in this base).
A: With the examples so far you might come to think that a matrix is diagonalizable iff it is a bijection. However it might also pose a hurdle, if the underlying field is not complete:
Consider the rotation matrix $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ on $\mathbb{R}^{2\times2}$:
It's eigenvectors are $\begin{pmatrix} 1 \\ \pm i \end{pmatrix}$, its eigenvalues $\mp i$, therefore it is not diagonalizable on $\mathbb{R}^{2\times2}$. 
Maybe wikipedia can give you some more insights: https://en.wikipedia.org/wiki/Diagonalizable_matrix#Examples
