When we say that a linear map $T:V\to V$ where $\dim(V)=n$ has a $n$ linear independent vectors so $T$ can be diagonalisable we refer not to every $n$ linear independent vectors, it must be the eigenvectors? as
$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has linear independent vectors but it is not diagonalisable
On
You're missing the key point:
$T:V \to V$ where $\dim V=n$ is diagonizable if and only if there are $n$ linearly independent vectors that are all eigenvectors of $T$.
Not just any vectors will do. Your example
$$ A=\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix} $$
has only one eigenvalue $\lambda=1$. Note all eigenvectors are non-zeor solutions to the equation $(A-\lambda I)v=0$ where $I$ is the $2\times 2$ identity matrix. Calculate the following: $$ A-1I=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix} $$
It should be clear that the only possible eigenvector has form $v=t\begin{bmatrix}1\\0\end{bmatrix}$ for some non-zero $t$. Thus it's impossible to find $2$ linearly indepndent eigenvectors for $A$
Recall that to diagonalize a n-by n matrix what we need is a basis of $n$ (independent) eigenvectors.
Note also that we don't need independent columns vectors (or rows) for a matrix to be diagonalizable, as for example
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
which is of course in diagonal form.
Refer also to Diagonalizable matrix.