Write down a linear operator $f:\mathbb{R}^4\to\mathbb{R}^4$ whose minimal polynomial is $m_f(t)=t^3-t^2$ Write down a linear operator $f: \mathbb{R}^4\to\mathbb{R}^4$ whose minimal polynomial is $m_f(t)=t^3-t^2$
I know that since in $\mathbb{R}^4$ we will have a characteristic polynomial $p_f(t)=t^3(t-1)$ or $p_f(t)=t^2(t-1)^2$. 
But we have to check $m_f(f)=0$ for our selected matrix which would involve a lot of matrix multiplication, I was wondering if there was an easier way to do this without having to plug our matrix into the minimal polynomial and writing pages of matrix multiplication for each test.
Or what would be an example of a matrix that I could use, and how did you decide on that matrix.
Any help and hints would be greatly appreciated.
For example I had:
$$A=\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$
I compute the characteristic polynomial:
$p_f(t)=det(A-tI)=det\begin{pmatrix}-t&0&0&0\\0&-t&0&0\\0&0&1-t&0\\0&0&0&1-t\end{pmatrix}=t^2(1-t)^2$, now how would I go about checking if it was minimal.
It would take ages to check and this was only a 3 mark question, there must be a quicker way, or a more obvious answer.
 A: The trick here is that a $k\times k$ matrix of the form
$$
A=\left(\begin{array}{ccccc}
\lambda&1&0&\cdots&0\\
0&\lambda&1&\cdots&0\\
0&0&\lambda&\cdots&0\\
\vdots&\vdots&\ddots&\ddots&\vdots\\
0&0&0&\cdots&\lambda\end{array}\right)
$$
with $\lambda$s on the diagonal and $1$s immediately above the diagonal has minimal polynomial $(t-\lambda)^k$.
To write down a $4\times4$ matrix with minimal polynomial $t^2(t-1)$ you need diagonal blocks for $\lambda=0$ and $\lambda=1$. The $\lambda=0$ block has to have size at least $2\times2$, but for $\lambda=1$ a $1\times1$ block is enough. That adds up to only $3\times3$, so you can add an extra $1\times1$ diagonal block belonging to either $\lambda=1$ or $\lambda=0$. Which? Up to you! If you know the characteristic polynomial that would help, but now you seem to have a choice.

For more information study Jordan canonical forms.
A: If the operator had to operate on $\mathbb{R}^3$ then the companion matrix $M$ of $m_f$ would do the trick, but this can be easily amended to an operator on  $\mathbb{R}^4$ using a block matrix. Indeed the matrix $\left(\begin{array}{cc} 1&0\\0&M\\          \end{array}\right)$
 works fine because 1 is a root of $m_f$. For the same reason $\left(\begin{array}{cc}
0&0\\
0&M\\                
\end{array}\right)$ also works.
