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Let us consider a subset $S$ of $M_4(\mathbb R)$ which has following form \begin{align*} \begin{pmatrix} 0 & * & 0 & * \\ 1 & * & 0 & * \\ 0 & * & 0 & * \\ 0 & * & 1 & * \end{pmatrix}, \end{align*} where $*$ can assume any real number. It is also clear for any monic $4^{th}$ degree real polynomial, we can at least find one realization in $S$ since the upper left block and lower right block can be considered as in companion form. Let $f: S \to \mathbb R^n$ be the map sending the coefficients of characteristic polynomial to $\mathbb R^n$.

My question is: suppose we have a square-free polynomial $p(t) = t^4 + a_3 t^3 + a_2 t^2 + a_1 t + a_0$, is $f^{-1}(a)$ a connected set in $M_n(\mathbb R)$ where $a=(a_3, a_2, a_1, a_0)$? If we let $C$ denote the companion form of $p(t)$, which is an element of $S$, then $$f^{-1}(a) = \{V C V^{-1}: V \in GL_4(\mathbb R), V C V^{-1} \text{ is in above form }\}.$$ I have a feeling we can first choose a realization in block form, \begin{align*} \begin{pmatrix} 0 & * & 0 & 0 \\ 1 & * & 0 & 0 \\ 0 & 0 & 0 & * \\ 0 & 0 & 1 & * \end{pmatrix}, \end{align*} and then continuously change everything in $f^{-1}(a)$ into this form without changing the characteristic polynomial. Indeed, if we consider a matrix \begin{align*} A = \begin{pmatrix} 0 & -b_1 & 0 & -c_1 \\ 1 & -b_2 & 0 & -c_2 \\ 0 & -b_3 & 0 & -c_3 \\ 0 & -b_4 & 1 & -c_4 \\ \end{pmatrix}, \end{align*} then the charateristic polynomial is \begin{align*} t^4 + (c_4+b_2)t^3 + (c_3 + b_1 + b_2c_4 - b_4 c_3)t^2 + (b_1 c_4 - b_3 c_3 + b_2 c_3 - b_4 c_1) t + (b_1 c_3 - b_3 c_1). \end{align*} I am thinking we should be able to continuously decrease $b_3, b_4, c_1, c_2$ to $0$ while keeping the polynomial unchanged by changing $b_1, b_2, c_3, c_4$ accordingly.

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  • $\begingroup$ The characteristic polynomial contains two typos. Here is the correct one: \begin{align*} t^4 + (c_4+b_2)t^3 + (c_3 + b_1 + b_2c_4 - b_4 c_2)t^2 + (b_1 c_4 - b_3 c_2 + b_2 c_3 - b_4 c_1) t + (b_1 c_3 - b_3 c_1). \end{align*} $\endgroup$ – Helmut Aug 12 '18 at 9:35
  • $\begingroup$ If $b_3>0$ then it is possible to reduce first $b_1$, then $b_2$ and $b_4$ to 0 and at last $b_3$ to 1 in order to obtain a companion matrix. I think the cases $c_1>0$, $c_3>0$ and $b_1>0$ can be treated similarly and, since the characteristic polynomial is square free these 4 entries cannot be all zero simultaneously. Maybe it is possible to extend this to a proof, but the cases of negative $b_3$ etc add another problem as one of them would have to be reduced to positive ones in a first step. $\endgroup$ – Helmut Aug 12 '18 at 9:53
  • $\begingroup$ Please ignore my previous comment. I made a mistake. $\endgroup$ – Helmut Aug 12 '18 at 13:59
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$\newcommand{\NN}{{\mathbb{N}}}\newcommand{\CC}{{\mathbb{C}}}\newcommand{\RR}{{\mathbb{R}}}\newcommand{\ra}{\rightarrow}\newcommand{\ds}{\displaystyle}$

The answer might be surprising: If the characteristic polynomial $p(t)$ has at least one real zero then the set $f^{-1}(p)$ is connected. If $p(t)$ has no real zeros than it consists of three connected components.

The starting point of my solution is the observation that the characteristic polynomial $p(t)= t^4 + a_3 t^3 + a_2 t^2 + a_1 t + a_0$ of $A$ in $f^{-1}(p)$, \begin{align*} A = \begin{pmatrix} 0 & -b_1 & 0 & -c_1 \\ 1 & -b_2 & 0 & -c_2 \\ 0 & -b_3 & 0 & -c_3 \\ 0 & -b_4 & 1 & -c_4 \\ \end{pmatrix}, \end{align*} is calculated as \begin{equation}\tag{1}\label{eq1} p(t)= (t^2+b_2t+b_1)(t^2+c_4t+c_3)-(b_4t+b_3)(c_2t+c_1). \end{equation} It is tempting to vary $b_4,b_3,c_2,c_1$ as parameters and determine $b_2,b_1,c_4,c_3$ from a real factorisation of the polynomial $p(t)+(b_4t+b_3)(c_2t+c_1)$. While such a factorisation always exists (not unique in general), it does not depend continuously upon the parameters, that is if we vary $b_4,b_3,c_2,c_1$ continuously on some path in $\mathbb R^4$, then it might be impossible to choose corresponding factorisations in a continuous way. This can be seen using the complex roots of $(t^2+b_2t+b_1)(t^2+c_4t+c_3)$, but it is not the question here.

Instead, I will use $b_1,b_2,b_3,b_4$ as parameters and determine $c_1,c_2,c_3,c_4$ by solving the linear equation (\ref{eq1}). This leads to a system of linear equations \begin{equation}\tag{2}\label{eq2} \begin{array}{rcrcrcrcl} &&&&&&c_4&=&a_3-b_2\\ &-&b_4\,c_2&+&c_3&-&b_2\,c_4&=&a_2-b_1\\ -b_4c_1&-&b_3\,c_2&+&b_2\,c_3&+&b_1\,c_4&=&a_1\\ -b_3\,c_1&&&+&b_1\,c_3&&&=&a_0. \end{array} \end{equation} Its determinant is $d(b_1,b_2,b_3,b_4)=b_3^2-b_2b_3b_4+b_1b_4^2$. If $b_4\neq0$ then it vanishes if and only if the two polynomials $t^2+b_2t+b_1$ and $b_4t+b_3$ have a common zero, namely $t=-b_3/b_4$. Whenever $d(b_1,b_2,b_3,b_4)\neq0$, the parameters $b_j$ determine $c_1,c_2,c_3,c_4$ uniquely. If the parameters vary continuously along some path in $\RR^4$ on which $d$ does not vanish, then so do the corresponding $c_j$.

Given a matrix $A$ in $f^{-1}(p)$, we will now try to reduce it within $f^{-1}(p)$ to the block diagonal form indicated by MyCindy2012. During the proof, it will also become clear that all such block diagonal matrices can be connected by paths within $f^{-1}(p)$. We have to consider several cases. The trivial ones that $b_3=b_4=0$ or $c_1=c_2=0$ are excluded in the sequel because $c_1,c_2$ or $b_3,b_4$, respectively, can be reduced to zero without changing anything else.

  1. If $b_4=0$ and $b_3\neq0$, then $d(b_1,b_2,b_3,b_4)\neq0$ whatever $b_1,b_2$ and hence they can be chosen arbitrarily and, by (\ref{eq2}), $c_1,c_2,c_3,c_4$ are uniquely determined. Therefore we can reduce $b_1,b_2$ to those values correponding to some real factorisation $p(t)= (t^2+b_2t+b_1)(t^2+c_4t+c_3)$ of $p$. By the uniqueness, we must then have $c_1=c_2=0$ and we reach the block diagonal form. Observe that in this case, we can reach {\em any} real factorisation of $p(t)$. Trivially the present case can be reached from any real factorisation of $p(t)$ ad therefore they can all be connected by paths within $f^{-1}(p)$. In the sequel we assume that $b_4\neq0$ tacitly.

  2. If $d(b_1,b_2,b_3,b_4)=b_3^2-b_2b_3b_4+b_1b_4^2 >0$ then we can first increase $b_1$ to a sufficiently large value, then reduce $b_2$ to $0$ and finally reduce $b_4$ to 0 without leaving the set $b_3^2-b_2b_3b_4+b_1b_4^2 >0$. Completing these $b_j$ by $c_k$ from system (\ref{eq2}), we obtain a path in the subset of $f^{-1}(p)$ on which $b_3^2-b_2b_3b_4+b_1b_4^2 >0$. As it leads to case 1, we are done.

  3. If $d(b_1,b_2,b_3,b_4)=0$, then $t^2+b_2t+b_1$ and $b_4t+b_3$ have common zero $z=-b_3/b_4$ and thus a common factor $t-z$. By (\ref{eq1}), $t-z$ must also be a factor of $p(t)$ and we can divide (\ref{eq1}) by $t-z$ to obtain \begin{equation}\tag{3}\label{eq3} q(t)=t^3+d_2t^2+d_1t+d_0=(t+h)(t^2+c_4t+c_3)-g_1t-g_2 \end{equation} with $d_0,d_1,d_2,g_1,g_2,h$ related to previous coefficients. We write $q=F(c_3,c_4,g_1,g_2,h)$.
    We now show that for any given $q$, the set $F^{-1}(q)$ is connected. It is convenient to assume that $d_2=0$. This is not a loss of generality as replacing $t$ by $t-\frac13d_2$ in (\ref{eq3}) leads to an equivalent situation. If we denote the zeroes of $t^2+c_4t+c_3$ by $z_1,z_2$ (Attention, . they might be conjugate complex) then for $s$, $0\leq s\leq 1$, the product $t^2+sc_4t+s^2c_3=(t-sz_1)(t-sz_2)$ and the product $(t+sh)(t^2+sc_4t+s^2c_3)$ has the zeroes $sh,sz_1,sz_2$. Putting $$g_1(s)t+g_2(s)=(t+sh)(t^2+sc_4t+s^2c_3)-q(t),$$ we obtain for each $s$ a decomposition (\ref{eq3}) in $F^{-1}(q)$. Therefore any decomposition (\ref{eq3}), $d_2=0$, can be reduced within $F^{-1}(q)$ to $q(t)=t\cdot t^2+d_1t+d_0$ by letting $s$ vary from 1 to 0. As a consequence, any two decompositions (\ref{eq3}) are connected within $F^{-1}(q)$.
    For our purpose, we reduce $g_1,g_2$ to 0. In the corresponding problem for our polynomial of degree 4, this means that we can reach $c_1=c_2=0$ and we are done.

  4. If $d(b_1,b_2,b_3,b_4)=b_3^2-b_2b_3b_4+b_1b_4^2 <0$ and $p$ has at least one real zero, say $t=z$, then we first reduce $b_1$ to a sufficiently large negative value, $b_2$ to 0 and then reduce $b_3$ to the value $b_3=-z\,b_4$. If $b_1$ is sufficiently negative, we remain in the set $b_3^2-b_2b_3b_4+b_1b_4^2 <0$ and can therefore complete with certain $c_j$ to elements of $f^{-1}(p)$. Now $p(t)$ and $b_4t+b_3$ have a common zero: $t=z$. By (\ref{eq1}), it must also be a zero of $t^2+c_4t+c_3$, since it is not a zero of $t^2+b_2t+b_1$. Therefore $t^2+c_4t+c_3$, $b_4t+b_3$ and $p(t)$ can be divided by $t-z$ and we obtain a problem for a polynomial of degree 3 analogous to the one in case 3. Again, we can reach the block diagonal form.

  5. If $d(b_1,b_2,b_3,b_4)=b_3^2-b_2b_3b_4+b_1b_4^2 <0$ and $p$ has no real zeroes then we cannot reach a block diagonal form. In fact, we cannot reach any point in $f^{-1}(p)$ where $d$ vanishes (and hence also no point where $d$ is positive). To show this, assume that we have decompositions \begin{equation} p(t)= (t^2+b_2(s)t+b_1(s))(t^2+c_4(s)t+c_3(s))-(b_4(s)t+b_3(s))(c_2(s)t+c_1(s)). \end{equation} with coefficients depending continuously upon $s$, $0\leq s\leq 1$, that $b_3(s)^2-b_2(s)b_3(s)b_4(s)+b_1(s)b_4(s)^2<0$ for $s<1$ whereas $b_3(1)^2-b_2(1)b_3(1)b_4(1)+b_1(1)b_4(1)^2=0$. We cannot have $b_4(1)\neq0$ because then, as shown in case 3, $p(t)$ has a linear term $t-z$ as a factor contradicting the assumption that it does not have real zeroes. Therefore $b_4(1)=b_3(1)=0$ and for $s=1,$ we have reached a real factorisation of $p(t)$. Hence the two quadratic factors have no real roots and therefore negative discriminant, in particular $b_2(1)^2-4b_1(1)<0$.
    On the other hand, for $s<1$ we must have $b_4(s)\neq0$ and the polynomial $t^2+b_2(s)t+b_1(s)$ has a negative value at $t=-b_3(s)/b_4(s)$. Therefore, it has positive discriminant: $b_2(s)^2-4b_1(s)>0$ for $s<1.$ By continuity, we must have $b_2(1)^2-4b_1(1)\geq0$ and have reached a contradiction.
    Observe that is this case, we must also have $c_1^2-b_2c_1c_2+b_1c_2^2<0$, because in (\ref{eq1}), $c_2t+c_1$ might replace the other linear term $b_4t+b_3$. If $c_1^2-b_2c_1c_2+b_1c_2^2\geq0$ then we can reach a block diagonal decomposition in $f^{-1}(c)$ -- as shown in case 2 and 3 -- but this is not possible. Similarly, we have $b_3^2-c_4b_3b_4+c_3b_4^2<0$, $c_1^2-c_4c_1c_2+c_3c_2^2<0$.
    We also show that the subset of $f^{-1}(p)$ on which $b_3^2-b_2b_3b_4+b_1b_4^2 <0$ has two connected components. Clearly, it must have at least two components, because $b_4$ cannot vanish on it. On the other hand from any starting point, we can first decrease $b_1$ to sufficiently large negative $b_1$ and then reduce $b_2$ and $b_3$ to 0 and $b_4$ to $1$ or $-1$. The corresponding $c_j$ are again uniquely determined from system (\ref{eq2}). Therefore the subset of $f^{-1}(p)$ on which $b_3^2-b_2b_3b_4+b_1b_4^2 <0$ has exactly two connected components in the present case and together with case 2, $f^{-1}(p)$ has exactly 3 connected components in the case that $p$ has no real zeroes.

Edit: 1. Observe that it is not necessary in this proof that $p(t)$ is squarefree.
2. In the complex domain. $f^{-1}(p)$ is always connected. $p(t)$ has always zeros, factorisations into quadratic factors can be continued as cntinuous functions and elimination of finitely many points of $\mathbb C$ still leaves a connected set.
3. For completeness, here is an example of a continuous family of monic real polynomials of degree 4 for which there does not exist a continuous real factorisation into quadratic factors. We define the polynomials by giving their zeros.
First for $s$ from $1$ downto $0$, the zeros are $\pm1\pm s\,i$, then for $a$ from $1$ downto $0$, they are $\pm1$ and $\pm s$, finally for $b$ from $0$ to $1$ they are $\pm1$ and $\pm bi$.
The factorisation must then be (except for ordering) $(t^2+2t+1+s^2)(t^2-2t+1+s^2)$ for $s$ from $1$ downto 0, then $(t^2+(1+a)t+a)(t^2-(1+a)t+a)$ for $a$ from 1 downto 0 and it jumps to $(t^2-1)(t^2+b^2)$ at this point.

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  • $\begingroup$ Very nice answer. I asked a generalized version of the question here math.stackexchange.com/questions/2889278/…. Please take a look if possible. $\endgroup$ – user1101010 Aug 20 '18 at 22:54
  • $\begingroup$ @user9527 In the case of two diagonal companion blocks, my answer can be generalised to some extent to blocks of any size. If the characteristic polynomial has enough zeros (the exact number is to be determined yet) then $f^{-1}(a)$ is connected. The exact number of connected components in all cases can probably also be determined, but this will be tedious (probably). In the cases of more than two diagonal companion blocks, I do not yet understand the condition generalising $b_3^2-b_2b_3b_4+b_1b_4^2=0$ enough. To be followed. $\endgroup$ – Helmut Aug 21 '18 at 10:34
  • $\begingroup$ Thanks. I would be more interested to see what polynomial would result in a connected preimage. Do you have any candidates in mind for me to gain intuition? For example, it seems to me $p(t) = t^n$ or $p(t) = (t-1)^n$ are likely to be special. But I am not sure. $\endgroup$ – user1101010 Aug 21 '18 at 17:14
  • $\begingroup$ @Helmut: Hi Helmut. I forgot how I understood your answer in the first place. But now I could not see: in case 1, why we can reach to any factorization if we reduce the matrix to a block diagonal form? More specifically, if we have two block realizations of the same polynomial, how do we connect them? $\endgroup$ – MyCindy2012 Oct 30 '18 at 7:21
  • $\begingroup$ @MyCindy2012: If one starts at a real factorisation $p(t)=(t^2+\tilde b_2 t+\tilde b_1)(t^2+\tilde c_4 t+\tilde c_3)$, it corresponds to block diagonal form: $\tilde b_3=\tilde b_4=\tilde c_1=\tilde c_2=0$. $p(t)$ does not change if we modify only $\tilde b_3$ to 1, say. As we keep $\tilde b_4=0$, we are in case 1 and can modify $\tilde b_1,\tilde b_2$ into the coefficients $\bar b_1,\bar b_2$ of any other factorisation $p(t)=(t^2+\bar b_2 t+\bar b_1)(t^2+\bar c_4 t+\bar c_3)$ along some path; the other coefficients are modified as indicated in case 1: We can reach any other factorisation. $\endgroup$ – Helmut Oct 30 '18 at 9:24

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