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Let $$A =\begin{pmatrix}0 & 0 & c \\1 & 0 & b \\ 0& 1 & a\end{pmatrix}$$ Show that the characteristic and minimal polynomials of $A$ are the same.

I have already computated the characteristic polynomial

$$p_A(x)=x^3-ax^2-bx-c$$

and I know from here that if I could show that the eigenspaces of $A$ all have dimension $1$, I would be done. The problem is that solving for the eigenvalues of this (very general) cubic equation is difficult (albeit possible), meaning it would be difficult to find bases for the eigenspaces.

A hint would be appreciated.

Rodrigo de Azevedo
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zz20s
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3 Answers3

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Compute: $$A^2 = \begin{pmatrix} 0 & c & ac \\ 0 & b & c + ab \\ 1 & a & b + a^2\end{pmatrix}.$$ So, we just need to show that $A^2, A, I$ are linearly independent. Clearly $A$ is not a multiple of $I$, so we just need to show there is no solution to the equation $$A^2 = pA + qI \iff \begin{pmatrix} 0 & c & ac \\ 0 & b & c + ab \\ 1 & a & b + a^2\end{pmatrix} = p\begin{pmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a\end{pmatrix} + q\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$ for $p$ and $q$. In particular, if you examine the entries in the left column, bottom row, we get $$1 = 0p + 0q,$$ which means there is indeed no solution. Hence $I, A, A^2$ are linearly independent, so no quadratic of $A$ will be equal to the $0$ matrix. Thus, the minimal polynomial must be (at least) a cubic, and equal to the characteristic polynomial

Theo Bendit
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  • Interesting! Can you elaborate on the sentence "$I, A, A^2$ are linearly independent, so no quadratic of $A$ will be equal to $0$"? – zz20s May 17 '19 at 00:44
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    Nice solution. Your argument can be rewritten as: If $c_1A^2+c_2A+c_3I_3=0_3$ then, looking at the first columns we get $$\begin{bmatrix} c_1 \c_2 \ c_3 \end{bmatrix}=\begin{bmatrix} 0 \ 0\0 \end{bmatrix}$$ – N. S. May 17 '19 at 00:45
  • @zz20s To say that $I, A, A^2$ are linearly dependent is to say that there are some scalars $p, q, r$, not all equal to $0$, such that $pA^2 + qA + rI = 0$. That is, there is some non-zero polynomial $f(x) = px^2 + qx + r$, of degree at most $2$, such that $f(A) = 0$. So, $I, A, A^2$ being independent means that there are no polynomials $f$ of degree less than $3$ (except the $0$ polynomial) such that $f(A) = 0$. Hence, the minimal polynomial must have degree at least $3$. – Theo Bendit May 17 '19 at 00:48
  • Ah, right, thank you! That makes sense. Is this a standard method for proving such a statement, or does it only work because of some property inherent to this matrix? – zz20s May 17 '19 at 00:54
  • It's a method that should work every time, provided you can solve the equations. If you're given an $n \times n$ matrix $A$ that you wish to show is diagonalisable, then this is equivalent to showing $I, A, A^2, \ldots, A^{n-1}$ are linearly independent. You can always do this mechanically, but sometimes it might mean solving a system of $n^2$ equations in $n$ variables! This matrix is particularly nice because the independence could be essentially read off three entries (cf N. S.'s comment). – Theo Bendit May 17 '19 at 01:01
  • Interesting. Thank you for your solution! – zz20s May 17 '19 at 01:03
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The form of $A$ has a special name: the companion matrix of the polynomial $p(x)=x^3-ax^2-bx-c$.

For the standard basis $e_1,e_2,e_3$, one finds that $Ae_1=e_2$, $Ae_2=e_3$, so $\{e_1,Ae_1,A^2e_1\}$ forms a basis.

The general context is the companion $n\times n$ matrix of the polynomial $$p(x)=x^n-c_{n-1}x^{n-1}-\cdots-c_1x-c_0.$$ A vector $v$ is said to be a cyclic vector for $A$ if the iterates by $A$ of $v$ for a basis for $R^n$. As others point out, this suffices to show that the minimal polynomial is the same as the characteristic polynomial.

user52817
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Assuming you know already that according to Cayley-Hamilton you have $p_A(A) = O_{3\times 3}$ you can also proceed as follows:

  • Let $e_1, e_2, e_3$ denote the canonical basis $\Rightarrow Ae_1=e_2, Ae_2 = e_3 \Rightarrow A^2e_1 = e_3$

Now, assume there is a polynomial $m(x)=x^2+ux+v$ such that $m(A) = O_{3\times 3}$.

Applying $m(A)$ to $e_1$ gives $$m(A)e_1 = A^2e_1 + uAe_1 + ve_1 = e_3 +ue_2 + ve_1 = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix} \mbox{ Contradiction!}$$ The linear combination cannot result in the zero vector as the coefficient of the basis vector $e_3$ is $1$.

trancelocation
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