I want to find the minimal polynomial $m(x)$ of the matrix below- no brute force allowed ( although it's feasible here)
$A=\begin{pmatrix}
1 & 0 & 0 & 0\\
1 & 1 & 0 & 0\\
2 & 1 & 2 & 0\\
2 & 1 & 1 & 2\\
\end{pmatrix}
.$
Here's my attempt:
I notice that $A$ has eigenvector $v_1 =\begin{pmatrix}
0 \\
0 \\
0\\
1
\end{pmatrix}$ with eigenvalue $2$. Restricted to the eigenspace of $v_1$, call it $E_1$, $A$ has minimal polynomial $m_1(x)=(x-2)$. $A$ acts on the quotient space $\mathbb{R}^4/E_1$ via the matrix
$A_1=\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
2 & 1 & 2
\end{pmatrix}$,
which has eigenvector $\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}$ with eigenvalue $2$ and eigenspace $E_2$, say. Restricted to $E_2$, $A$ has minimal polynomial $m_2(x)=(x-2)$. $A_1$ acts on the $\mathbb{R}^3/E_2$ via the matrix
$A_2= \begin{pmatrix}
1 & 0\\
1 & 1
\end{pmatrix}$,
which clearly has minimal polynomial $m_2(x)=(x-1)^2$.
Combining all these results $m(x)= (x-1)^2(x-2)^2$.
Is this correct? Is there a better way to do this? Many thanks!
Best Answer
Your "proof" is non-sense. To find the minimal polynomial, you must first find $A$-invariant spaces. For example, $span(e_4)$ is invariant but $span(e_1,e_2,e_3)$ is not.
Here, it suffices to show that the minimal polynomial has degree $4$. Do as follows
i) Randomly choose a vector $x$.
ii) Considering the invariant space generated by $x$, we obtain the matrix $M=[x,Ax,A^2x,A^3x]$. If $\det(M)\not= 0$, then $x$ generates the whole space and the minimal polynomial of $A$ is the characteristic polynomial of $A$.
That works with $x=[1,1,1,1]^T$ (which is not very random; try another vector...).
EDIT. Answer to the OP.
For the minimal polynomial, you can't separate some characteristic space into two parts. For example, here, you may consider whole characteristic space $\ker((A-2I)^2)$ which is $span(e_3,e_4)$ -an invariant subspace-. The matrix of the restriction is $\begin{pmatrix}2&0\\1&2\end{pmatrix}$ and its minimal polynomial is $(x-2)^2$.
Now you consider $\mathbb{R^4}/span(e_3,e_4)$; the associated matrix (using the canonical basis) is $\begin{pmatrix}1&0\\1&1\end{pmatrix}$; its minimal polynomial is $(x-1)^2$ and we are done.
In particular, the matrix
$\begin{pmatrix}1&0&0&0\\a&1&0&0\\b&c&2&0\\d&e&f&2\end{pmatrix}$ has $(x-1)^2(x-2)^2$ as minimal polynomial iff $af\not= 0$.