Quadratic form as a homogenous polynomial












1












$begingroup$


Let $Q(v)=v'Av$, where$v=begin{pmatrix}x&y&z&wend{pmatrix} A=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&1\0&0&1&0end{pmatrix}$, then does there exist an invertible matrix $P$ such that $Q(Pv)=x^2+y^2-zw$? Note that all terms are in $mathbb{R}$. I think we should use orthogonal diagonalization of $A$. But, the term $zw$ is intriguing






linear-algebra quadratic-forms







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  • $begingroup$
    Are you working over the reals, the complex numbers, or some other field?
    $endgroup$
    – Servaes
    Dec 12 '18 at 11:24










  • $begingroup$
    @Servaes Oh! sorry, I am working over reals
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 14:36
















1












$begingroup$


Let $Q(v)=v'Av$, where$v=begin{pmatrix}x&y&z&wend{pmatrix} A=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&1\0&0&1&0end{pmatrix}$, then does there exist an invertible matrix $P$ such that $Q(Pv)=x^2+y^2-zw$? Note that all terms are in $mathbb{R}$. I think we should use orthogonal diagonalization of $A$. But, the term $zw$ is intriguing






linear-algebra quadratic-forms







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you working over the reals, the complex numbers, or some other field?
    $endgroup$
    – Servaes
    Dec 12 '18 at 11:24










  • $begingroup$
    @Servaes Oh! sorry, I am working over reals
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 14:36














1












1








1





$begingroup$


Let $Q(v)=v'Av$, where$v=begin{pmatrix}x&y&z&wend{pmatrix} A=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&1\0&0&1&0end{pmatrix}$, then does there exist an invertible matrix $P$ such that $Q(Pv)=x^2+y^2-zw$? Note that all terms are in $mathbb{R}$. I think we should use orthogonal diagonalization of $A$. But, the term $zw$ is intriguing






linear-algebra quadratic-forms







share|cite|improve this question











$endgroup$




Let $Q(v)=v'Av$, where$v=begin{pmatrix}x&y&z&wend{pmatrix} A=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&1\0&0&1&0end{pmatrix}$, then does there exist an invertible matrix $P$ such that $Q(Pv)=x^2+y^2-zw$? Note that all terms are in $mathbb{R}$. I think we should use orthogonal diagonalization of $A$. But, the term $zw$ is intriguing






linear-algebra quadratic-forms




linear-algebra quadratic-forms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 14:37







vidyarthi

















asked Dec 12 '18 at 10:47









vidyarthividyarthi

2,9641832




2,9641832












  • $begingroup$
    Are you working over the reals, the complex numbers, or some other field?
    $endgroup$
    – Servaes
    Dec 12 '18 at 11:24










  • $begingroup$
    @Servaes Oh! sorry, I am working over reals
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 14:36


















  • $begingroup$
    Are you working over the reals, the complex numbers, or some other field?
    $endgroup$
    – Servaes
    Dec 12 '18 at 11:24










  • $begingroup$
    @Servaes Oh! sorry, I am working over reals
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 14:36
















$begingroup$
Are you working over the reals, the complex numbers, or some other field?
$endgroup$
– Servaes
Dec 12 '18 at 11:24




$begingroup$
Are you working over the reals, the complex numbers, or some other field?
$endgroup$
– Servaes
Dec 12 '18 at 11:24












$begingroup$
@Servaes Oh! sorry, I am working over reals
$endgroup$
– vidyarthi
Dec 12 '18 at 14:36




$begingroup$
@Servaes Oh! sorry, I am working over reals
$endgroup$
– vidyarthi
Dec 12 '18 at 14:36










1 Answer
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$begingroup$

Note that $Q(Pv)=v^{top}P^{top}APv$, so one approach is to find $P$ such that
$$P^{top}AP=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&-frac{1}{2}\0&0&-frac{1}{2}&0end{pmatrix}.$$
As the first two rows and columns of $P^{top}AP$ are the same as those of $A$, it makes sense to have $P$ leave those alone, i.e. to take $P$ of the form
$$P=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&a&b\0&0&c&dend{pmatrix}.$$
This reduces the problem to a problem on $2times2$-matrices, which is very manageable without any theory. We get the system of equations
begin{eqnarray*}
ac+ba&=&0,\
c^2+ad&=&-frac{1}{2},\
b^2+ad&=&-frac{1}{2},\
cd+bd&=&0
end{eqnarray*}
which immediately implies $b^2=c^2$, and $b+c=0$ or $a=d=0$. This immediately shows the easy solution $a=d=0$ and $b=c=frac{1}{sqrt{2}}i$. The remaining (family of) solutions is not hard to determine from the system above.






share|cite|improve this answer









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  • $begingroup$
    great and succinct answer
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 11:37











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1 Answer
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1 Answer
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2












$begingroup$

Note that $Q(Pv)=v^{top}P^{top}APv$, so one approach is to find $P$ such that
$$P^{top}AP=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&-frac{1}{2}\0&0&-frac{1}{2}&0end{pmatrix}.$$
As the first two rows and columns of $P^{top}AP$ are the same as those of $A$, it makes sense to have $P$ leave those alone, i.e. to take $P$ of the form
$$P=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&a&b\0&0&c&dend{pmatrix}.$$
This reduces the problem to a problem on $2times2$-matrices, which is very manageable without any theory. We get the system of equations
begin{eqnarray*}
ac+ba&=&0,\
c^2+ad&=&-frac{1}{2},\
b^2+ad&=&-frac{1}{2},\
cd+bd&=&0
end{eqnarray*}
which immediately implies $b^2=c^2$, and $b+c=0$ or $a=d=0$. This immediately shows the easy solution $a=d=0$ and $b=c=frac{1}{sqrt{2}}i$. The remaining (family of) solutions is not hard to determine from the system above.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    great and succinct answer
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 11:37
















2












$begingroup$

Note that $Q(Pv)=v^{top}P^{top}APv$, so one approach is to find $P$ such that
$$P^{top}AP=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&-frac{1}{2}\0&0&-frac{1}{2}&0end{pmatrix}.$$
As the first two rows and columns of $P^{top}AP$ are the same as those of $A$, it makes sense to have $P$ leave those alone, i.e. to take $P$ of the form
$$P=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&a&b\0&0&c&dend{pmatrix}.$$
This reduces the problem to a problem on $2times2$-matrices, which is very manageable without any theory. We get the system of equations
begin{eqnarray*}
ac+ba&=&0,\
c^2+ad&=&-frac{1}{2},\
b^2+ad&=&-frac{1}{2},\
cd+bd&=&0
end{eqnarray*}
which immediately implies $b^2=c^2$, and $b+c=0$ or $a=d=0$. This immediately shows the easy solution $a=d=0$ and $b=c=frac{1}{sqrt{2}}i$. The remaining (family of) solutions is not hard to determine from the system above.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    great and succinct answer
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 11:37














2












2








2





$begingroup$

Note that $Q(Pv)=v^{top}P^{top}APv$, so one approach is to find $P$ such that
$$P^{top}AP=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&-frac{1}{2}\0&0&-frac{1}{2}&0end{pmatrix}.$$
As the first two rows and columns of $P^{top}AP$ are the same as those of $A$, it makes sense to have $P$ leave those alone, i.e. to take $P$ of the form
$$P=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&a&b\0&0&c&dend{pmatrix}.$$
This reduces the problem to a problem on $2times2$-matrices, which is very manageable without any theory. We get the system of equations
begin{eqnarray*}
ac+ba&=&0,\
c^2+ad&=&-frac{1}{2},\
b^2+ad&=&-frac{1}{2},\
cd+bd&=&0
end{eqnarray*}
which immediately implies $b^2=c^2$, and $b+c=0$ or $a=d=0$. This immediately shows the easy solution $a=d=0$ and $b=c=frac{1}{sqrt{2}}i$. The remaining (family of) solutions is not hard to determine from the system above.






share|cite|improve this answer









$endgroup$



Note that $Q(Pv)=v^{top}P^{top}APv$, so one approach is to find $P$ such that
$$P^{top}AP=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&0&-frac{1}{2}\0&0&-frac{1}{2}&0end{pmatrix}.$$
As the first two rows and columns of $P^{top}AP$ are the same as those of $A$, it makes sense to have $P$ leave those alone, i.e. to take $P$ of the form
$$P=begin{pmatrix}1&0&0&0\0&1&0&0\0&0&a&b\0&0&c&dend{pmatrix}.$$
This reduces the problem to a problem on $2times2$-matrices, which is very manageable without any theory. We get the system of equations
begin{eqnarray*}
ac+ba&=&0,\
c^2+ad&=&-frac{1}{2},\
b^2+ad&=&-frac{1}{2},\
cd+bd&=&0
end{eqnarray*}
which immediately implies $b^2=c^2$, and $b+c=0$ or $a=d=0$. This immediately shows the easy solution $a=d=0$ and $b=c=frac{1}{sqrt{2}}i$. The remaining (family of) solutions is not hard to determine from the system above.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 11:28









ServaesServaes

24.9k33893




24.9k33893












  • $begingroup$
    great and succinct answer
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 11:37


















  • $begingroup$
    great and succinct answer
    $endgroup$
    – vidyarthi
    Dec 12 '18 at 11:37
















$begingroup$
great and succinct answer
$endgroup$
– vidyarthi
Dec 12 '18 at 11:37




$begingroup$
great and succinct answer
$endgroup$
– vidyarthi
Dec 12 '18 at 11:37


















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