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Expanding

$p(x)=(ax-b)(cx+d)$

we get

$acx^2+(ad-bc)x-bd$.

Notice the determinant of the matrix

$\begin{pmatrix} a & b \\ c & d \end{pmatrix} $

is $ad-bc$ exactly like the constant of $x$ in the polynomial expansion.

So I searched links between the equation and the matrix determinant. I only found the solution of the equation $p(x)=0$ to be $x \in \{b/a, -d/c\}$. Then I found numerous other things, but none of them were simple enough to be considered worthy remembering.

My question: What are the correlations between the polynomials and determinants of the matrices with entries being the coefficients of the polynomial?

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    $\begingroup$ If $A$ is a $2\times 2$ matrix, then its characteristic equation is $\lambda^2 - \operatorname{Tr}(A)\lambda + \operatorname{Det}(A)$. This helps a lot when trying to classify the behavior of dynamical systems. $\endgroup$ – Antonio Vargas Mar 3 '13 at 17:35
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It is not entirely clear to me what you are asking, but you might be interested in knowing that any monic polynomial (say with real coefficients, but any ring will do) occurs as the characteristic polynomial of some matrix with coefficients in $\mathbb R$ (or the same field the coefficients of the polynomial are coming from). It's a nice exercise to find the matrix (hint: it's more straightforward than one might initially think).

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  • $\begingroup$ It's not clear how this "explains" the OP's observations. Could you please provide further details. $\endgroup$ – Math Gems Mar 3 '13 at 15:58
  • $\begingroup$ @MathGems if I had understood wht OP was trying to ask I would not have stated that I don't quite understand what is being asked. I was merely trying to give something useful, taking a stab in the dark. $\endgroup$ – Ittay Weiss Mar 3 '13 at 19:29
  • $\begingroup$ Ah, I see, it's not an answer but, rather, a comment about something tangentially related. I was scratching my head, puzzled, mistakenly thinking that this answer might be a hint about something closely connected to the problem at hand. $\endgroup$ – Math Gems Mar 3 '13 at 19:40
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    $\begingroup$ Presumbaly you refer to the companion matrix. While I do know some ways to relate it to the OP's problem, I do not see any explicit mention of any particular relationship in your answer. That's what I was hoping you might explain further. By the way, questioners frequently accept answers without (fully) understanding them. $\endgroup$ – Math Gems Mar 3 '13 at 20:02
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    $\begingroup$ "By the way, questioners frequently accept answers without (fully) understanding them." Is that a bug, or a feature? $\endgroup$ – Gerry Myerson Mar 3 '13 at 22:58
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A very important link between polynomials and determinants is this: if $A$ is a square matrix, $I$ is the identity matrix, and $x$ is an indeterminate, then $\det A$ is the constant term of the polynomial $\det(A-xI)$.

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  • $\begingroup$ It's not clear how this "explains" the OP's observations. Could you please provide further details. $\endgroup$ – Math Gems Mar 3 '13 at 16:01
  • $\begingroup$ OP didn't ask for an explanation of observations. OP asked, "What are the correlations between the polynomials and determinants of the matrices with entries being the coefficients of the polynomial?" and I gave one example. $\endgroup$ – Gerry Myerson Mar 3 '13 at 22:57
  • $\begingroup$ Ah, I see now. You are interpreting the OP's question very generally, e.g. for different types of polynomials, but I read it as referring only to the specific polynomial he discussed. It's not clear which was intended. $\endgroup$ – Math Gems Mar 3 '13 at 23:03
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Conceptually, one nice way to "explain" this is as follows. We know from high-school that for a polynomial $\rm\: x^n - \color{#C00}{c_{n-1}}x^{n-1}\!+\,\cdots+c_0,\:$ the coefficient $\rm\,\color{#C00}{c_{n-1}}\,$ equals the sum of the roots. Hence $\rm\: \ell(f)\, :=\, \color{#C00}{c_{n-1}}$ satisfies a $\rm\color{blue}{logarithmic}$ law $\rm\ \ell(fg)\, =\, \ell(f)+\ell(g)\ $ (as is also easily verified directly). Your observation may be interpreted as a variation on this logarithmic property of root sums.

Define $\rm\,\ f\,\left[\begin{array}{cc}\rm a&\rm b\\ \rm c&\rm d\end{array}\right]\, :=\ \ell((ax\!-\!b)(cx\!+\!d))\, :=\, $ coefficient of $\rm\,x\,$ (the "generic" value of $\ell)$

  • $\rm\,f\,$ is multilinear: $\, $ True, essentially, by the above $\rm\color{blue}{logarithmic}$ property of $\:\ell$.

  • $\rm\,f\,$ is alternating: $\rm\, \ f\,\left[\begin{array}{cc}\rm a&\rm b\\ \rm a&\rm b\end{array}\right] =\ \ell((ax\!-\!b)(ax\!+\!b))\, =\, \ell(a^2x^2\!+\color{#C00}0\cdot x-\!b^2)\, =\, \color{#C00}0$

  • $\rm\,f\,$ is normalized: $\rm\ f(I) = 1\:$ by $\rm\: f\,\left[\begin{array}{cc}\rm 1&\rm 0\\ \rm 0&\rm 1\end{array}\right] =\ \ell(1\!\cdot\! x\!-\!0)(0\!\cdot\! x\!+\!1))\, =\, \ell(\color{#C00}1\cdot x) = \color{#C00}1$

However, as is well-known, $\rm\:det\,$ (determinant) is the unique solution of the above equations. Therefore, by uniqueness, we deduce that $\rm\ f = det.$

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  • $\begingroup$ Using \rm everywhere makes your math hard to read. $\endgroup$ – Antonio Vargas Mar 3 '13 at 17:31
  • $\begingroup$ Thank you for your explanation. $\endgroup$ – Dávid Natingga Mar 3 '13 at 21:54

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