<s>[INST] How do I have to use this assumption? 
Let be $V$ a vector space over $\mathbb{R}$ with dimension 4. Let be
$T$ a linear operator in $V$. In case that a base $\mathcal{B}$ of $V$
exists, such that the associated matrix of $T$ relative to the basis
$\mathcal{B}$ is: $$ \left [ T \right ]_{\mathcal{B}}=\begin{pmatrix}
 1 & a & 0 & 0\\  0 & 1 & b & 0\\  0 & 0 & 1 & c\\  0 & 0 & 0 & 1
 \end{pmatrix} $$ with $a^{2}+b^{2}+c^{2} \neq 0$, determine if T is
diagonalizable.

My solution:
Let be $A=\left [ T \right ]_{\mathcal{B}}$
First we compute $det(A-tI)$ to find the eigenvalues of A, that is
\begin{align*}
det(A-tI)=(1-t)^{4}
\end{align*}
Then, we have an unique eigenvalue $\lambda=1$ with algebraic multiplicity $\mu_{1}=4$. Now, we have to find the eigenspace of the eigenvalue $\lambda=1$, so we have to solve this:
\begin{align*}
\begin{pmatrix}
0 & a & 0 & 0\\ 
0 & 0 & b & 0\\ 
0 & 0 & 0 & c\\ 
0 & 0 & 0 & 0 
\end{pmatrix}
\begin{pmatrix}
x_1\\ 
x_2\\ 
x_3\\ 
x_4
\end{pmatrix}=\begin{pmatrix}
0\\ 
0\\ 
0\\ 
0
\end{pmatrix}
\end{align*}
Thus,
\begin{align*}
a \cdot x_2&=0& & &x_2=0 & &\\
b \cdot x_3&=0& &\Longrightarrow &x_3=0 &  &\wedge& &x_1=r\\
c \cdot x_4&=0& & &x_3=0 & &
\end{align*}
Therefore, the eigenspace is:
\begin{align*}
S_{1}=\left \{ r \begin{pmatrix}
1\\ 
0\\ 
0\\ 
0
\end{pmatrix} : \forall r \in F \right \}
\end{align*}
So, $\dim(S_{1})=1 \neq \mu_{1}=4$ $\Longrightarrow$ $T$ is not diagonalizable.
My questions are:

*

*How do I have to use the assumption that $a^2+b^2+c^2\neq 0$?

*Is my solution correct?

(I have already asked this question here but in this post I put my complete solution and a new question about the assumption. I would really really appreciate your help.)
 [/INST] Note that $a^2 + b^2 + c^2 = 0$, for $a, b, c \in \mathbb{R}$, only holds if $a = b = c = 0$.
Your error comes in this step:
\begin{align*}
a \cdot x_2&=0& & &x_2=0 & &\\
b \cdot x_3&=0& &\Longrightarrow &x_3=0 &  &\wedge& &x_1=r\\
c \cdot x_4&=0& & &x_3=0 & &
\end{align*}
Because in this step, you've assumed that $a, b, c \neq 0$ (you also made a typo, since the last line should be $x_4 = 0$, not $x_3$). If one of the parameters is 0, then the corresponding $x$ value is actually free (e.g. if $a = 0$, then $x_2 = s \in \mathbb{R}$).
So the dimension of the eigenspace of $A$ is equal to one more than the number of parameters that is equal to zero. As long as at least one of $a, b, c$ is non-zero, your argument is correct and $T$ is not diagonalisable. If $a = b = c = 0$, then the eigenspace of $A$ is $\mathbb{R}^4$ and $T$ is diagonalisable (and, in fact, if $a = b = c = 0$ then $A$ is the identity matrix which is not just diagonalisable but already diagonal).
</s>