minimal polynomial of a matrix with some unknown entries

Question

Question is to prove that :

characteristic and minimal polynomial of $ \left( \begin{array}{cccc} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.

what i have done so far is :

characteristic polynomial of a matrix $A$ is given by $\det(A-xI)$

in case of $A= \left( \begin{array}{cccc} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{array} \right)$ we have $\det(A-xI)=\det\left( \begin{array}{cccc} -x & 0 & c \\ 1 & -x & b \\ 0 & 1 & a-x \end{array} \right)=-(x^3-ax^2-bx-c)$

So, i have got the characteristic polynomial as $x^3-ax^2-bx-c$.

Now, the problem is how do i find minimal polynomial.

As $a,b,c$ are arbitrary, I can not factorize $x^3-ax^2-bx-c$ so as to see which factor gives me minimal polynomial.

I am confused.

please suggest me some hint.

EDIT : This is just after Mr.Will Jagyy's hint :

I have $A= \left( \begin{array}{cccc} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{array} \right)$ then, $A^2= \left( \begin{array}{cccc} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{array} \right)\left( \begin{array}{cccc} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{array} \right)=\left( \begin{array}{cccc} 0 & c & ac \\ 0 & b & c+ab \\ 1 & a & b+a^2 \end{array} \right)$

Now,

$A^2+rA+sI=\left( \begin{array}{cccc} 0 & c & ac \\ 0 & b & c+ab \\ 1 & a & b+a^2 \end{array} \right)+r\left( \begin{array}{cccc} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{array} \right)+s\left( \begin{array}{cccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc} s & c & * \\ r & b+s & * \\ 1& * & * \end{array} \right)$

As element of $3^{rd}$ row $1^{st}$ column is $1$ in above matrix, this can never be $0$

i.e., $A^2+rA+sI$ can never be $0$.

Now, $A+rI=\left( \begin{array}{cccc} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{array} \right)+r\left( \begin{array}{cccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc} r & *& * \\ \color{magenta}{1} & * & * \\ * & \color{magenta}{1} & * \end{array} \right)\neq 0$ if $r\neq 0$

Thus, $A^2+rA+sI\neq 0$ and $A+rI\neq 0$ for any $r,s$.

Thus, minimal polynomial for $A$ can not be of order less than $3$.

Thus, minimal polynomial for $A$ has to be $x^3-ax^2-bx-c$.

I have written this just to make sure i have tried in correct way as i can not write this in a comment.

I would be thankful if there is any other way to proceed further.. Thank you :)

Answer 1

It is very easy to see what the minimal polynomial is for this matrix, even without knowing let alone factoring the characteristic polynomial. The argument even works for matrices of this type (called companion matrices) of arbitrary size $n$. Since $A\cdot e_i=e_{i+1}$ for $0<i<n$, where $e_i$ is the standard basis vector, it is clear that $(c_0I+c_1A+c_2A^2+\cdots+c_{n-1}A^{n-1})\cdot e_1=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n$, which is obviously nonzero unless all the coefficients $c_i$ are zero. Therefore no nonzero polynomial of degree${}<n$ in$~A$ has $e_1$ in its kernel, and a fortiori no such polynomial can be the zero matrix. Therefore the minimal polynomial of$~A$ must have degree at least$~n$, so that it must coincide with the characteristic polynomial by Cayley-Hamilton.

You computed the characteristic polynomial in your example, but it would be a bit harder (though doable) for larger companion matrices. For the minimal polynomial it is easy by reasoning similarly to above. If the entries down the last column are $a_1,a_2,\ldots,a_n$, then $A\cdot e_n=a_1e_1+a_2e_2+\cdots+a_ne_n$. Now the minimal polynomial of$~A$ must be of the form $\mu=c_0+c_1X+\cdots+c_{n-1}X^{n-1}+X^n$, and computing as above one has $\mu[A]\cdot e_1=c_0e_1+c_1e_2+\cdots+c_{n-1}e_n+A^n\cdot e_1$, which must be zero; given that $A^n\cdot e_1=A\cdot e_n$ has the value just computed, it must be that $c_i+a_i=0$ for all $i$, so $c_i=-a_i$ fo all $i$ and hence $\mu=-a_0-a_1X+\cdots-a_{n-1}X^{n-1}+X^n$. You can conclude (using Cayley-Hamilton) that this is also the characteristic polynomial of$~A$. You can also prove this by direct calculation, either by induction on$~n$ and development of the determinant by the first row, or (without induction) by development of the determinant by the final column.

Answer 2

You don't need to factor anything. Your characteristic polynomial, call it $f(x) = x^3 - a x^2 - b x - c,$ satisfies $f(A) = 0,$ meaning $$ A^3 - a A^2 - b A - c I = 0 $$ as matrices. This is Hamilton-Cayley. Actually, given your question, you should carefully check that with actual matrices. What are $A^2$ and $A^3?$

So, there is a (monic) minimal polynomial, call it $m(x).$ Maybe $m(x) = f(x).$ But maybe $m$ is quadratic or even linear. So, carefully write out the matrix expression $$ A^2 + r A + s I $$ and see if any choice of $r,s$ results in the zero matrix. Then, check $$ A + t I $$ to see if any choice of $t$ results in the zero matrix.

If the minimal polynomial is not quadratic or linear, then it is cubic.

Meanwhile, i collected a bunch of equivalent conditions at Given a matrix, is there always another matrix which commutes with it? which includes an overall answer to your question.

Answer 3

Here's how I would argue the point: clearly the minimal polynomial $m_A(x)$ of $A$, where

$A = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix}, \tag{1}$

must satisfy $\deg m_A(x) \le 3$, since the characteristic poynomial $p_A(x)$ of $A$ satisfies $\deg p_A(x) = 3$. Furthermore, we also have $m_A(x) \mid p_A(x)$. Thus $p_A(x) = q(x)m_A(x)$ for some polynomial $q(x)$. Now if $\deg m_A(x) = 3$, we must have $\deg q(x) = 0$ since $\deg p_A(x) = \deg q(x) + \deg m_A(x)$. This, combined with the fact that $m_A(x), p_A(x)$ are both monic, leads to the conclusion that $q(x) = 1$ and hence $m_A(x) = p_A(x)$; we are done in this case. Now suppose that $\deg m_A(x) = 1$; then $m_A(x)$ must be of the form $m_A(x) = x - d$ for some constant $d$. Then since $m_A(A) = 0$, we obtain $A = dI$, where $I$ is the $3 \times 3$ identity matrix; then $A$ is diagonal; but this contradicts the fact that $A_{21} = A_{32} = 1$ for any choice of $a, b, c$; thus the possibility that $\deg m_A(x) = 1$ may be excluded. The case $\deg m_A(x) = 2$ remains; here we have $m_A(x) = x^2 - dx - e$ for some $d, e$. Then $m_A(A) = 0$ implies $A^2 = dA + eI$. We compute $A^2$:

$A^2 = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix} \begin{bmatrix} 0 & 0 & c\\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix} =\begin{bmatrix} 0 & c & ac \\ 0 & b & ab + c \\ 1 & a & a^2 + b \end{bmatrix}, \tag{2}$

and observe that $A^2_{31} = 1$ for all $a, b, c$; on the other hand, inspection of (1) shows that, for any $d, e$,

$(dA + e)_{31} = 0; \tag{3}$

we thus conclude that $\deg m_A(x) = 2$ is impossible. This leaves us with $\deg m_A(x) = 3$, and as we have seen, $m_A(x) = p_A(x)$ in this situation. QED.

Additional Observations: Some curiosities of the matrix $A$: Suppose we write $A$ in the form

$A = A_1 + N, \tag{4}$

where

$A_1 = \begin{bmatrix} 0 & 0 & c \\ 0 & 0 & b \\ 0 & 0 & a \end{bmatrix}, \tag{5}$

and

$N = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}; \tag{6}$

then we have

$A_1^2 = \begin{bmatrix} 0 & 0 & ac \\ 0 & 0 & ab \\ 0 & 0 & a^2 \end{bmatrix}, \tag{7}$

and

$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}; \tag{8}$

then

$A^2 = (A_1 + N)^2 = A_1^2 + A_1 N + N A_1 + N^2, \tag{9}$

with

$A_1 N = \begin{bmatrix} 0 & c & 0 \\ 0 & b & 0 \\ 0 & a & 0 \end{bmatrix} \tag{10}$

and

$N A_1 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & c \\ 0 & 0 & b \end{bmatrix} \tag{11}$

and

$N^3 = 0, \tag{12}$

i.e. $N$ is nilpotent of degree $3$; furthermore, (7) shows that the third column of $A_1$, $(c, b, a)^T$, is in fact an eigenvector of $A$ with eigenvalue $a$; thus

$A_1^m = \begin{bmatrix} 0 & 0 & a^{m - 1}c \\ 0 & 0 & a^{m - 1}b \\ 0 & 0 & a^m \end{bmatrix} \tag{13}$

readily follows; in contrast to (12), $A_1^m \ne 0$ as long as $a$ does not vanish.

Not sure where all this leads, but it seems that the peculiar form of $N$, $N^2$ etc. has something to do with the above argument that $\deg m_A(x) \ne 2$. My guts tell me there is a generalization waiting to appear. Like William Buther Yeats' "rough beast" (see his poem The Second Coming), something is "slouching towards Bethlehem to be born."

END: Additional Observations: Some curiosities of the matrix $A$.

Hope at least some of this spiel helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 4

The matrix of that form is called the "companion matrix"

Let $\{e_1,e_2,e_3\}$ be the standard basis for $\mathbb{F}^3$.

Then we see that $Ae_1=e_2, Ae_2=e_3, Ae_3=ce_1+be_2+ae_3$

Give a $\mathbb{F}[x]$-module structure on $\mathbb{F}^3$ by $x$-action defined as $A$-multiplication.

Denote that $\mathbb{F}[x]$-module structure as $M^A$.

Then $M^A\simeq \mathbb{F}[x]/(x^3-ax^2-bx-c)$ as $\mathbb{F}[x]$-module.

The RHS clearly shows that $A$ has minimal polynomial $x^3-ax^2-bx-c$.