Necessary and sufficient condition for quadratic forms
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Consider the quadratic form $Q(u,v)=au^2 + 2buv +cv^2$. Upon completing the square we will obtain $$Q(u,v)=aleft(u+frac{bv}{a}right)^2+left(c-frac{b^2}{a}right)v^2,$$
we assume that $ane 0$.
There is a theorem which says that
A set of necessary and sufficient conditions that $Q(u,v)$ be positive for all nontrivial $u$ and $v$ is that
$$a>0, text{ } left|begin{matrix}
a & b\
b & c
end{matrix}right|>0$$
However, what if $a>0$ but $left(c-frac{b^2}{a}right)<0$? Then either $c$ is negative or $c<frac{b^2}{a}$. Why wouldn't this satisfy $Q(u,v)$ as being positive?
This is also equivalent to
$$ac-b^2>0 text{ (1)}$$
$$ac-b^2<0 text{ (2)}$$
with (1) implying that $c>0$ whenever $a>0$ but $ac$ can also be positive when $a<0$ and $c<0$ and $ac>b^2$. Likewise, (2) can also be negative when $a,c>0$ and $ac<b^2$.
What is it that I'm missing here?
linear-algebra optimization quadratic-forms
asked Nov 22 at 18:30
sequence
4,19331135
add a comment |
up vote
0
down vote
favorite
Consider the quadratic form $Q(u,v)=au^2 + 2buv +cv^2$. Upon completing the square we will obtain $$Q(u,v)=aleft(u+frac{bv}{a}right)^2+left(c-frac{b^2}{a}right)v^2,$$
we assume that $ane 0$.
There is a theorem which says that
A set of necessary and sufficient conditions that $Q(u,v)$ be positive for all nontrivial $u$ and $v$ is that
$$a>0, text{ } left|begin{matrix}
a & b\
b & c
end{matrix}right|>0$$
However, what if $a>0$ but $left(c-frac{b^2}{a}right)<0$? Then either $c$ is negative or $c<frac{b^2}{a}$. Why wouldn't this satisfy $Q(u,v)$ as being positive?
This is also equivalent to
$$ac-b^2>0 text{ (1)}$$
$$ac-b^2<0 text{ (2)}$$
with (1) implying that $c>0$ whenever $a>0$ but $ac$ can also be positive when $a<0$ and $c<0$ and $ac>b^2$. Likewise, (2) can also be negative when $a,c>0$ and $ac<b^2$.
What is it that I'm missing here?
linear-algebra optimization quadratic-forms
asked Nov 22 at 18:30
sequence
4,19331135
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the quadratic form $Q(u,v)=au^2 + 2buv +cv^2$. Upon completing the square we will obtain $$Q(u,v)=aleft(u+frac{bv}{a}right)^2+left(c-frac{b^2}{a}right)v^2,$$
we assume that $ane 0$.
There is a theorem which says that
A set of necessary and sufficient conditions that $Q(u,v)$ be positive for all nontrivial $u$ and $v$ is that
$$a>0, text{ } left|begin{matrix}
a & b\
b & c
end{matrix}right|>0$$
However, what if $a>0$ but $left(c-frac{b^2}{a}right)<0$? Then either $c$ is negative or $c<frac{b^2}{a}$. Why wouldn't this satisfy $Q(u,v)$ as being positive?
This is also equivalent to
$$ac-b^2>0 text{ (1)}$$
$$ac-b^2<0 text{ (2)}$$
with (1) implying that $c>0$ whenever $a>0$ but $ac$ can also be positive when $a<0$ and $c<0$ and $ac>b^2$. Likewise, (2) can also be negative when $a,c>0$ and $ac<b^2$.
What is it that I'm missing here?
linear-algebra optimization quadratic-forms
asked Nov 22 at 18:30
sequence
4,19331135
Consider the quadratic form $Q(u,v)=au^2 + 2buv +cv^2$. Upon completing the square we will obtain $$Q(u,v)=aleft(u+frac{bv}{a}right)^2+left(c-frac{b^2}{a}right)v^2,$$
we assume that $ane 0$.
There is a theorem which says that
A set of necessary and sufficient conditions that $Q(u,v)$ be positive for all nontrivial $u$ and $v$ is that
$$a>0, text{ } left|begin{matrix}
a & b\
b & c
end{matrix}right|>0$$
However, what if $a>0$ but $left(c-frac{b^2}{a}right)<0$? Then either $c$ is negative or $c<frac{b^2}{a}$. Why wouldn't this satisfy $Q(u,v)$ as being positive?
This is also equivalent to
$$ac-b^2>0 text{ (1)}$$
$$ac-b^2<0 text{ (2)}$$
with (1) implying that $c>0$ whenever $a>0$ but $ac$ can also be positive when $a<0$ and $c<0$ and $ac>b^2$. Likewise, (2) can also be negative when $a,c>0$ and $ac<b^2$.
What is it that I'm missing here?
linear-algebra optimization quadratic-forms
linear-algebra optimization quadratic-forms
asked Nov 22 at 18:30
sequence
4,19331135
asked Nov 22 at 18:30
sequence
4,19331135
asked Nov 22 at 18:30
sequence
4,19331135
asked Nov 22 at 18:30
sequence
4,19331135
asked Nov 22 at 18:30
sequence
4,19331135
4,19331135
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
when $a > 0$ but $c - frac {b^2}{a} < 0,$ test with pair
$$ u = -b, ; ; v = a . $$
Then
$$ a u^2 + 2 b u v + c v^2 = a b^2 - 2 b b a + c a^2 = ca^2 - a b^2 = a(ca-b^2) < 0 $$
answered Nov 22 at 19:25
Will Jagy
101k598198
add a comment |
up vote
0
down vote
Note indeed that the condition
$$left(c-frac{b^2}{a}right)<0 iff ac-b^2<0$$
is the opposite of
$$left|begin{matrix}
a & b\
b & c
end{matrix}right|>0 iff ac-b^2>0$$
answered Nov 22 at 18:40
gimusi
92.7k94495
Yes, but why can't $a>0$ in the first case if $c<frac{b^2}{a}$?
– sequence
Nov 22 at 18:45
@sequence Yes it can be but in that case (a=1, b=2, c=1) the quadratic forms is not positive definite.
– gimusi
Nov 22 at 18:50
Yes, but it is negative definite. The theorem above says that $a$ is necessarily negative for the negative definite case. But it doesn't have to be, or does it? @gimusi
– sequence
Nov 22 at 18:53
@sequence No in that case it is indefinite since in some "directions" $Q>0$ and in others $Q<0$. Indeed in that case the associated matrix has $lambda_1>0$ and $lambda_2<0$.
– gimusi
Nov 22 at 18:56
$a,b,c,d$ are assumed to be fixed constants. @gimusi
– sequence
Nov 22 at 18:58
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
when $a > 0$ but $c - frac {b^2}{a} < 0,$ test with pair
$$ u = -b, ; ; v = a . $$
Then
$$ a u^2 + 2 b u v + c v^2 = a b^2 - 2 b b a + c a^2 = ca^2 - a b^2 = a(ca-b^2) < 0 $$
answered Nov 22 at 19:25
Will Jagy
101k598198
add a comment |
up vote
1
down vote
accepted
when $a > 0$ but $c - frac {b^2}{a} < 0,$ test with pair
$$ u = -b, ; ; v = a . $$
Then
$$ a u^2 + 2 b u v + c v^2 = a b^2 - 2 b b a + c a^2 = ca^2 - a b^2 = a(ca-b^2) < 0 $$
answered Nov 22 at 19:25
Will Jagy
101k598198
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
when $a > 0$ but $c - frac {b^2}{a} < 0,$ test with pair
$$ u = -b, ; ; v = a . $$
Then
$$ a u^2 + 2 b u v + c v^2 = a b^2 - 2 b b a + c a^2 = ca^2 - a b^2 = a(ca-b^2) < 0 $$
answered Nov 22 at 19:25
Will Jagy
101k598198
when $a > 0$ but $c - frac {b^2}{a} < 0,$ test with pair
$$ u = -b, ; ; v = a . $$
Then
$$ a u^2 + 2 b u v + c v^2 = a b^2 - 2 b b a + c a^2 = ca^2 - a b^2 = a(ca-b^2) < 0 $$
answered Nov 22 at 19:25
Will Jagy
101k598198
answered Nov 22 at 19:25
Will Jagy
101k598198
answered Nov 22 at 19:25
Will Jagy
101k598198
answered Nov 22 at 19:25
Will Jagy
101k598198
101k598198
add a comment |
add a comment |
up vote
0
down vote
Note indeed that the condition
$$left(c-frac{b^2}{a}right)<0 iff ac-b^2<0$$
is the opposite of
$$left|begin{matrix}
a & b\
b & c
end{matrix}right|>0 iff ac-b^2>0$$
answered Nov 22 at 18:40
gimusi
92.7k94495
Yes, but why can't $a>0$ in the first case if $c<frac{b^2}{a}$?
– sequence
Nov 22 at 18:45
@sequence Yes it can be but in that case (a=1, b=2, c=1) the quadratic forms is not positive definite.
– gimusi
Nov 22 at 18:50
Yes, but it is negative definite. The theorem above says that $a$ is necessarily negative for the negative definite case. But it doesn't have to be, or does it? @gimusi
– sequence
Nov 22 at 18:53
@sequence No in that case it is indefinite since in some "directions" $Q>0$ and in others $Q<0$. Indeed in that case the associated matrix has $lambda_1>0$ and $lambda_2<0$.
– gimusi
Nov 22 at 18:56
$a,b,c,d$ are assumed to be fixed constants. @gimusi
– sequence
Nov 22 at 18:58
|
show 1 more comment
up vote
0
down vote
Note indeed that the condition
$$left(c-frac{b^2}{a}right)<0 iff ac-b^2<0$$
is the opposite of
$$left|begin{matrix}
a & b\
b & c
end{matrix}right|>0 iff ac-b^2>0$$
answered Nov 22 at 18:40
gimusi
92.7k94495
Yes, but why can't $a>0$ in the first case if $c<frac{b^2}{a}$?
– sequence
Nov 22 at 18:45
@sequence Yes it can be but in that case (a=1, b=2, c=1) the quadratic forms is not positive definite.
– gimusi
Nov 22 at 18:50
Yes, but it is negative definite. The theorem above says that $a$ is necessarily negative for the negative definite case. But it doesn't have to be, or does it? @gimusi
– sequence
Nov 22 at 18:53
@sequence No in that case it is indefinite since in some "directions" $Q>0$ and in others $Q<0$. Indeed in that case the associated matrix has $lambda_1>0$ and $lambda_2<0$.
– gimusi
Nov 22 at 18:56
$a,b,c,d$ are assumed to be fixed constants. @gimusi
– sequence
Nov 22 at 18:58
|
show 1 more comment
up vote
0
down vote
up vote
0
down vote
Note indeed that the condition
$$left(c-frac{b^2}{a}right)<0 iff ac-b^2<0$$
is the opposite of
$$left|begin{matrix}
a & b\
b & c
end{matrix}right|>0 iff ac-b^2>0$$
answered Nov 22 at 18:40
gimusi
92.7k94495
Note indeed that the condition
$$left(c-frac{b^2}{a}right)<0 iff ac-b^2<0$$
is the opposite of
$$left|begin{matrix}
a & b\
b & c
end{matrix}right|>0 iff ac-b^2>0$$
answered Nov 22 at 18:40
gimusi
92.7k94495
answered Nov 22 at 18:40
gimusi
92.7k94495
answered Nov 22 at 18:40
gimusi
92.7k94495
answered Nov 22 at 18:40
gimusi
92.7k94495
92.7k94495
Yes, but why can't $a>0$ in the first case if $c<frac{b^2}{a}$?
– sequence
Nov 22 at 18:45
@sequence Yes it can be but in that case (a=1, b=2, c=1) the quadratic forms is not positive definite.
– gimusi
Nov 22 at 18:50
Yes, but it is negative definite. The theorem above says that $a$ is necessarily negative for the negative definite case. But it doesn't have to be, or does it? @gimusi
– sequence
Nov 22 at 18:53
@sequence No in that case it is indefinite since in some "directions" $Q>0$ and in others $Q<0$. Indeed in that case the associated matrix has $lambda_1>0$ and $lambda_2<0$.
– gimusi
Nov 22 at 18:56
$a,b,c,d$ are assumed to be fixed constants. @gimusi
– sequence
Nov 22 at 18:58
|
show 1 more comment
Yes, but why can't $a>0$ in the first case if $c<frac{b^2}{a}$?
– sequence
Nov 22 at 18:45
@sequence Yes it can be but in that case (a=1, b=2, c=1) the quadratic forms is not positive definite.
– gimusi
Nov 22 at 18:50
Yes, but it is negative definite. The theorem above says that $a$ is necessarily negative for the negative definite case. But it doesn't have to be, or does it? @gimusi
– sequence
Nov 22 at 18:53
@sequence No in that case it is indefinite since in some "directions" $Q>0$ and in others $Q<0$. Indeed in that case the associated matrix has $lambda_1>0$ and $lambda_2<0$.
– gimusi
Nov 22 at 18:56
$a,b,c,d$ are assumed to be fixed constants. @gimusi
– sequence
Nov 22 at 18:58
Yes, but why can't $a>0$ in the first case if $c<frac{b^2}{a}$?
– sequence
Nov 22 at 18:45
Yes, but why can't $a>0$ in the first case if $c<frac{b^2}{a}$?
– sequence
Nov 22 at 18:45
@sequence Yes it can be but in that case (a=1, b=2, c=1) the quadratic forms is not positive definite.
– gimusi
Nov 22 at 18:50
@sequence Yes it can be but in that case (a=1, b=2, c=1) the quadratic forms is not positive definite.
– gimusi
Nov 22 at 18:50
Yes, but it is negative definite. The theorem above says that $a$ is necessarily negative for the negative definite case. But it doesn't have to be, or does it? @gimusi
– sequence
Nov 22 at 18:53
Yes, but it is negative definite. The theorem above says that $a$ is necessarily negative for the negative definite case. But it doesn't have to be, or does it? @gimusi
– sequence
Nov 22 at 18:53
@sequence No in that case it is indefinite since in some "directions" $Q>0$ and in others $Q<0$. Indeed in that case the associated matrix has $lambda_1>0$ and $lambda_2<0$.
– gimusi
Nov 22 at 18:56
@sequence No in that case it is indefinite since in some "directions" $Q>0$ and in others $Q<0$. Indeed in that case the associated matrix has $lambda_1>0$ and $lambda_2<0$.
– gimusi
Nov 22 at 18:56
$a,b,c,d$ are assumed to be fixed constants. @gimusi
– sequence
Nov 22 at 18:58
$a,b,c,d$ are assumed to be fixed constants. @gimusi
– sequence
Nov 22 at 18:58
|
show 1 more comment
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