Find a $3\times 3$ matrix whose minimal polynomial is $x^2$.
My try:
Since a characteristic polynomial and a minimal polynomial have the same roots ,so the characteristic polynomial must be $x^3$ since $0$ is the only characteristic value of multiplicity $2$.
So the matrix $A$ must be of the form \begin{bmatrix} 0 & 0 & 0\\ b & 0 & 0 \\ c & a & 0 \end{bmatrix}
Since the minimal polynomial is $x^2$ so rank $A=2$,so we must have a non-zero minor of order $2$ .Hence we should have $a\neq 0,b\neq 0;a,b,c\in \mathbb R$ .
Is the solution correct?Please suggest edits if required.
Best Answer
As you said, you know you're looking for an operator with characteristic polynomial $x^3$.
Since the minimal polynomial divides the characteristic polynomial, this operator has only two possible minimal polynomials: $x^2$ and $x$.
Therefore, you're looking for an operator $X$ with characteristic polynomial $x^3$ that satisfies $X^2 = 0$, but not $X = 0$.