Polynomial and determinant relationship?
$begingroup$
It can be proven that
If $$P(x)=ax^3+bx^2+cx+d$$ has two roots being opposite of each other
i.e. $alpha$ and $-alpha$, then $$ad=bc$$
This instantly made me think of a zero determinant of the general $2times2$ matrix. I cannot think of any obvious reason as to why this might be the case. Is this really just a coincidence, or is there more to it than this?
polynomials determinant
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
$endgroup$
add a comment |
$begingroup$
It can be proven that
If $$P(x)=ax^3+bx^2+cx+d$$ has two roots being opposite of each other
i.e. $alpha$ and $-alpha$, then $$ad=bc$$
This instantly made me think of a zero determinant of the general $2times2$ matrix. I cannot think of any obvious reason as to why this might be the case. Is this really just a coincidence, or is there more to it than this?
polynomials determinant
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
$endgroup$
add a comment |
$begingroup$
It can be proven that
If $$P(x)=ax^3+bx^2+cx+d$$ has two roots being opposite of each other
i.e. $alpha$ and $-alpha$, then $$ad=bc$$
This instantly made me think of a zero determinant of the general $2times2$ matrix. I cannot think of any obvious reason as to why this might be the case. Is this really just a coincidence, or is there more to it than this?
polynomials determinant
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
$endgroup$
It can be proven that
If $$P(x)=ax^3+bx^2+cx+d$$ has two roots being opposite of each other
i.e. $alpha$ and $-alpha$, then $$ad=bc$$
This instantly made me think of a zero determinant of the general $2times2$ matrix. I cannot think of any obvious reason as to why this might be the case. Is this really just a coincidence, or is there more to it than this?
polynomials determinant
polynomials determinant
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
asked Dec 11 '18 at 13:43
TrogdorTrogdor
5,12861843
5,12861843
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $beta$ be the third root. Thenbegin{align}P(x)&=a(x-alpha)(x+alpha)(x-beta)\&=a(x^3-beta x^2-alpha^2x+alpha^2beta)end{align}and therefore…
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
$endgroup$
add a comment |
$begingroup$
If $alpha=0$ then we must have $c=d=0$ so the result is obvious. If $alphaneq 0$, note that the matrix
$$
begin{bmatrix}
a&b\
c&d
end{bmatrix}
$$
sends both $begin{bmatrix}pmalpha\1end{bmatrix}$ (which are linearly independent) to the subspace $leftlanglebegin{bmatrix}1\-alpha^2end{bmatrix}rightrangle$. Hence it must have zero determinant, i.e., $ad-bc=0$.
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
votes
$begingroup$
Let $beta$ be the third root. Thenbegin{align}P(x)&=a(x-alpha)(x+alpha)(x-beta)\&=a(x^3-beta x^2-alpha^2x+alpha^2beta)end{align}and therefore…
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
$endgroup$
add a comment |
$begingroup$
Let $beta$ be the third root. Thenbegin{align}P(x)&=a(x-alpha)(x+alpha)(x-beta)\&=a(x^3-beta x^2-alpha^2x+alpha^2beta)end{align}and therefore…
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
$endgroup$
add a comment |
$begingroup$
Let $beta$ be the third root. Thenbegin{align}P(x)&=a(x-alpha)(x+alpha)(x-beta)\&=a(x^3-beta x^2-alpha^2x+alpha^2beta)end{align}and therefore…
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
$endgroup$
Let $beta$ be the third root. Thenbegin{align}P(x)&=a(x-alpha)(x+alpha)(x-beta)\&=a(x^3-beta x^2-alpha^2x+alpha^2beta)end{align}and therefore…
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
edited Dec 11 '18 at 19:25
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
answered Dec 11 '18 at 13:47
José Carlos SantosJosé Carlos Santos
158k22126229
158k22126229
add a comment |
add a comment |
$begingroup$
If $alpha=0$ then we must have $c=d=0$ so the result is obvious. If $alphaneq 0$, note that the matrix
$$
begin{bmatrix}
a&b\
c&d
end{bmatrix}
$$
sends both $begin{bmatrix}pmalpha\1end{bmatrix}$ (which are linearly independent) to the subspace $leftlanglebegin{bmatrix}1\-alpha^2end{bmatrix}rightrangle$. Hence it must have zero determinant, i.e., $ad-bc=0$.
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
$endgroup$
add a comment |
$begingroup$
If $alpha=0$ then we must have $c=d=0$ so the result is obvious. If $alphaneq 0$, note that the matrix
$$
begin{bmatrix}
a&b\
c&d
end{bmatrix}
$$
sends both $begin{bmatrix}pmalpha\1end{bmatrix}$ (which are linearly independent) to the subspace $leftlanglebegin{bmatrix}1\-alpha^2end{bmatrix}rightrangle$. Hence it must have zero determinant, i.e., $ad-bc=0$.
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
$endgroup$
add a comment |
$begingroup$
If $alpha=0$ then we must have $c=d=0$ so the result is obvious. If $alphaneq 0$, note that the matrix
$$
begin{bmatrix}
a&b\
c&d
end{bmatrix}
$$
sends both $begin{bmatrix}pmalpha\1end{bmatrix}$ (which are linearly independent) to the subspace $leftlanglebegin{bmatrix}1\-alpha^2end{bmatrix}rightrangle$. Hence it must have zero determinant, i.e., $ad-bc=0$.
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
$endgroup$
If $alpha=0$ then we must have $c=d=0$ so the result is obvious. If $alphaneq 0$, note that the matrix
$$
begin{bmatrix}
a&b\
c&d
end{bmatrix}
$$
sends both $begin{bmatrix}pmalpha\1end{bmatrix}$ (which are linearly independent) to the subspace $leftlanglebegin{bmatrix}1\-alpha^2end{bmatrix}rightrangle$. Hence it must have zero determinant, i.e., $ad-bc=0$.
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
answered Dec 11 '18 at 13:58
user10354138user10354138
7,3872925
7,3872925
add a comment |
add a comment |
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