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.)
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).