Is $A = \left( {\matrix{ 3 & 0 \cr 1 & 3 \cr } } \right)$ diagonalizable? Prove it using characteristic/minimal polynomial
$$f_A(x)=(x-3)^2$$
I think the minimal polynomial is $m_A = x-3 \ne f_A(x)$ and therefore, $A$ isn't diagonalizable. Am I right?
On
that $A = \pmatrix{3 & 0\cr 1 & 3}$ can be show without invoking minimal polynomial of this matrix. we will use the following two facts:
(a) $A$ is diagonalizable if and only if $A - 3I,$
(b) eigenvalues of a lower triangular matrix are its diagonal elements,
here is the argument that $A$is not diagonalizable. suppose it is diagonalizable. then so is $A - 3I,$ but the eigenvalues of $A - 3I$ are $0$ twice. this implies $A-3I = \pmatrix{0 & 0 \cr 1 & 0} = S \pmatrix{0 & 0 \cr 0 & 0} S^{-1} = \pmatrix{0 & 0 \cr 0 & 0}.$ this is a contradiction, therefore neither $A$ nor $A-3I$ is diagonalizable.
The polynomial $x-3$ is not the minimal polynomial, since $$ A-3I_2=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix} $$ which is not the zero matrix.
The minimal polynomial is indeed $(x-3)^2$ as you can easily verify. Since this polynomial doesn't have pairwise distinct roots, the matrix is not diagonalizable.
The fact that $m_A\ne f_A$ doesn't imply the matrix is non diagonalizable: think to the identity matrix $I_2$, which has minimal polynomial $x-1$ and characteristic polynomial $(x-1)^2$, but is certainly diagonalizable.
Another way to see $A$ is not diagonalizable is by computing the geometric multiplicity of the eigenvalue $3$, which is $1$, while its algebraic multiplicity is $2$.