Let $A=\begin{pmatrix}0 & a & c \\b & 0 & c \\ b& -a & 0 \\\end{pmatrix}$ . Investigate if the matrix A is diagonalizable if
a)$ A \in \mathbb{R}^{3 \times 3}$
b)$ A \in \mathbb{C}^{3 \times 3}$
I've made some progress for a): First of all I found the charachteristic polynomial:$f(x)=x^3+x(ac-ab-ac)$ and so it has 3 solutions ,$x_0=0$, $x_1,2=\pm \sqrt{b(a+c)-ac}$
$DimV_A(x_0)=1$ ,because $x_0$ is a simple solution (meaning algebraic dimension is 0).So if we want the matrix to be diagonalizable we need $DimV_A(x_1)=DimV_A(x_2) \Rightarrow rank(A-x_1I)=rank(A-x_2I)=2$.I would continue by making up some relationships to find a b and c.
This is where I'm getting a little insecure. There must be a more elegant way to show a) is either true or false.
If someone could help me with this or at least provide me with a hint ,it would be appreciated!
Best Answer
A sufficient but not necessary condition for a matrix to be diagonalizable is that her characteristic polynomial is a product of distinct linear factors, so as you mentioned in the comments, if $b(a+c)>ac$, the matrix is diagonalizable over $\mathbb{R}$ and $\mathbb{C}$.
If $b(a+c)=ac$, the characteristic polynomial has only one root with multiplicity 3, so our former condition fails, but it does not follows that the matrix is not diagonalizable.
A sufficient and necessary condition for a matrix to be diagonalizable is that it's minimal polynomial is a product of distinct linear factors. since in the case where $b(a+c)=ac$ the characteristic polynomial of the matrix $A$ is $x^3$, and because the minimal polynomial divides the characteristic polynomial, we can conclude that the minimal polynomial is of the form $x^m$ for $1\leq m\leq3$. Since $A\neq 0$ we can conclude that $m\geq 2$ so in that case $A$ is not diagonalizable over $\mathbb{R}$ nor $\mathbb{C}$.
In the cae where $b(a+c)<ac$, the characteristic polynomial of $A$ has an irreducible factor over $\mathbb{R}$. Since the minimal polynomial has each irreducible factor of the characteristic polynomial as a factor (might be with less multiplicity), we can conclude that the minimal polynomial of $A$ is not a product of distinct linear factors over $\mathbb{R}$, so $A$ is not diagonalizable over $\mathbb{R}$ in this case.
However, in this case the characteristic polynomial is a product of distinct linear factors over $\mathbb{C}$, so in this case $A$ is diagonalizable over $\mathbb{C}$