Conditioning a linear operator in order to be diagonalizable.

Question

Let $T: \mathbb{R}^{4} \to \mathbb{R}^{4} $ be a linear operator such $$T(x_{1},x_{2},x_{3},x_{4})=(0,ax_{1},bx_{2},cx_{3}).$$

Find the conditions over the numbers $a,b$ and $c$ which make $T$ diagonalizable.

I try to analyze the behavior of $T$ by calculating $[T]_{\beta}$ when $\beta$ is the canonical basis for $\mathbb{R}^{4}$ but this is \begin{bmatrix}0&0&0&0\\a&0&0&0\\0&b&0&0\\0&0&c&0\end{bmatrix} So the characteristical polynomial is zero since $det([T]_{\beta})=0$ so which values for $a,b$ and $c$ work ? Thanks

Answer 1

Let $A = [T]_\beta$. Then the characteristic polynomial of $A$ is $$ \det(A - t I) = \det \begin{bmatrix} -t & 0 & 0 & 0\\ a & -t & 0 & 0\\ 0 & b & -t & 0\\ 0 & 0 & c & -t \end{bmatrix} = t^4 $$ so $0$ is an eigenvalue of algebraic multiplicity $4$. In order for $A$ to be diagonalizble, this means that the eigenspace of $\lambda = 0$, i.e., the null space of $A$, must have dimension $4$. This can only occur when $A$ is the zero matrix, that is, $a = b = c = 0$.

Answer 2

See here, that is:

a matrix is diagonalizable if and only if its minimal polynomial factors into distinct linear factors.

We have: $A\ne 0 \qquad A^2 \ne 0 \qquad A^3 \ne 0 \qquad A^4=0$

so the minimal polynomial of the matrix $A$ is $A^4$ that has four identical factors.