Finding minimal polynomial of lower triangular matrix

Question

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!

Answer 1

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 $af\not= 0$.

Answer 2

Yes the minmal polynomial is $(x-1)^2 (x-2)^2$. Another way to get that:

Since the matrix is lower triangular, its characteristic polynomial is $(x-1)^2 (x-2)^2$, and the minimal polynomial must divide this.
Note that both $A-I$ and $A-2I$ have rank $3$ and null space of dimension $1$. Since the dimension of the null space of a product of matrices is at most the sum of the dimensions of their null spaces, all four factors (two $(x-1)$'s and two $(x-2)$'s are needed to get a null space with dimension $4$.