positive definite problem

**transgalactic** · October 1st 2011, 02:50 AM

3)
A)
there is a lenear transformation T:V->V in inner product
it is given that $T(T^{*}-3I)+2I=0$ . prove that T is positive definite and find every posible
eigenvalue of T
?
B)
we define $q:R^{4}->R q(x_{1},x_{2},x_{3},x_{4})=2x_{1}x_{4}-6x_{2}x_{3}$
write the formula as a sum of squares,find the rank and signature of q
?
regarding A:
the book definition is:
a matrices which is hermitic A^{*}=A and for every vector and k of our choosing $(u,u)=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha k_{i}\overline{k_{j}}>0$
i dont know how to use it here
regarding B :
$q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}$
in the books definition a signature is \sigma=\pi-\mu
$\pi$ is the number of positive members=2
$\mu$ is the number of negative members =4
i got $\sigma=-2$
??
in order to find the rank i need to find representative matrices in
$x'=x_{1}+x_{4}$
$y'=x_{2}-3x_{3}$
and the rest stays the same
i got here 5 member in power 2
but my matrices is 4x4 i cant have 5 members in the diagonal of 4x4 matrices
??

**Deveno** · October 2nd 2011, 07:39 AM

for part B: note that: $2x_1x_4 = (\frac{x_1+x_4}{\sqrt{2}})^2 - (\frac{x_1-x_4}{\sqrt{2}})^2$

see if you can derive a similar expression for $-6x_2x_3$ . this should give you 2 positive terms, and 2 negative terms, for a total of 4.

**transgalactic** · October 2nd 2011, 08:40 AM

Originally Posted by transgalactic

3)
A)
there is a lenear transformation T:V->V in inner product
it is given that $T(T^{*}-3I)+2I=0$ . prove that T is positive definite and find every posible
eigenvalue of T
?
B)
we define $q:R^{4}->R q(x_{1},x_{2},x_{3},x_{4})=2x_{1}x_{4}-6x_{2}x_{3}$
write the formula as a sum of squares,find the rank and signature of q
?
regarding A:
the book definition is:
a matrices which is hermitic A^{*}=A and for every vector and k of our choosing $(u,u)=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha k_{i}\overline{k_{j}}>0$
i dont know how to use it here
regarding B :
$q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}$
in the books definition a signature is \sigma=\pi-\mu
$\pi$ is the number of positive members=2
$\mu$ is the number of negative members =4
i got $\sigma=-2$
??
in order to find the rank i need to find representative matrices in
$x'=x_{1}+x_{4}$
$y'=x_{2}-3x_{3}$
and the rest stays the same
i got here 5 member in power 2
but my matrices is 4x4 i cant have 5 members in the diagonal of 4x4 matrices
??

why this result is wrong i see that 5 members are a problem
$q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}$
but i completed the goal of transforming it into squared members only.
whats wrong with how i have done it?

**Deveno** · October 2nd 2011, 09:21 AM

think of it like "a change of variables". the set of variables {u1 = x1+x4, u2 = -x1, u3 = -x4, u4 = x2-3x3, u5 = -x2, u6 = -9x3} is not linearly independent,

whereas {u1 = x1+x4, u2 = x1-x4, u3 = x2+x3, u4 = x2-x3} is linearly independent.

**transgalactic** · October 2nd 2011, 11:24 AM

i used a method in my book called lagrange method.
it tells to complete the members into the (a+b)^2 formula

i see that my variables are dependant but i coulnt think of another way of solving it

is there a general method of solving this type of questions without getting a dependant set of variables in the end
?

**Deveno** · October 2nd 2011, 11:47 AM

look at post #2 again....

**transgalactic** · October 2nd 2011, 01:13 PM

you have showed there the end product
of 4 variables which are independant.

but i have an algorithm(lagrange method) of completing to (a+b)^2
and it brang me to a dead end solution

**Deveno** · October 2nd 2011, 10:38 PM

show the steps you took in your method, perhaps you over-looked something....

**transgalactic** · October 2nd 2011, 10:53 PM

Originally Posted by transgalactic

3)
A)
there is a lenear transformation T:V->V in inner product
it is given that $T(T^{*}-3I)+2I=0$ . prove that T is positive definite and find every posible
eigenvalue of T
?
B)
we define $q:R^{4}->R q(x_{1},x_{2},x_{3},x_{4})=2x_{1}x_{4}-6x_{2}x_{3}$
write the formula as a sum of squares,find the rank and signature of q
?
regarding A:
the book definition is:
a matrices which is hermitic A^{*}=A and for every vector and k of our choosing $(u,u)=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha k_{i}\overline{k_{j}}>0$
i dont know how to use it here
regarding B :
$q(x_{1},x_{2},x_{3},x_{4})=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}+(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}$
in the books definition a signature is \sigma=\pi-\mu
$\pi$ is the number of positive members=2
$\mu$ is the number of negative members =4
i got $\sigma=-2$
??
in order to find the rank i need to find representative matrices in
$x'=x_{1}+x_{4}$
$y'=x_{2}-3x_{3}$
and the rest stays the same
i got here 5 member in power 2
but my matrices is 4x4 i cant have 5 members in the diagonal of 4x4 matrices
??

i need to transform each one of these two members $2x_{1}x_{4}-6x_{2}x_{3}$ into to (a+b)^2 or
(a-b)^2
so i thought ,how do i get 2x_{1}x_{4} to be $(x_{1}+x_{4})^2$ .
and i just saw that i should substract the squares to get there.
$2x_{1}x_{4}=(x_{1}+x_{4})^{2}-x_{1}^{2}-x_{4}^{2}$
same thing for $6x_{2}x_{3}$
$6x_{2}x_{3}=(x_{2}-3x_{3})^{2}-x_{2}^{2}-9x_{3}^{2}$

**Deveno** · October 2nd 2011, 11:26 PM

is -x1^2 - x2^2 of either form (a+b)^2 or (a-b)^2? no, it is not.

however: (a+b)^2 - (a-b)^2 = 4ab, that is:

ab = ((a+b)/2)^2 - ((a-b)/2)^2 and these terms are of the required form (using a' = a/2, b' = b/2).

if you have a term like:

kab, then you use a' = a√k/2, b' = b√k/2.

your way gves you too many variables to be able to come up with a 4x4 matrix.

the goal is to replace each variable x1,x2,x3,x4 with some other variables u1,u2,u3,u4,

so that q(u1,u2,u3,u4) = ±(c1u1)^2 ± (c2u2)^2 ± (c3u3)^3 ± (c4u4)^2, so that the matrix for q

in the variables (u1,u2,u3,u4) is diag(±c1,±c2,±c3,±c4).

**transgalactic** · October 3rd 2011, 12:52 AM

Originally Posted by Deveno

is -x1^2 - x2^2 of either form (a+b)^2 or (a-b)^2? no, it is not.

however: (a+b)^2 - (a-b)^2 = 4ab, that is:

ab = ((a+b)/2)^2 - ((a-b)/2)^2 and these terms are of the required form (using a' = a/2, b' = b/2).

if you have a term like:

kab, then you use a' = a√k/2, b' = b√k/2.

your way gves you too many variables to be able to come up with a 4x4 matrix.

the goal is to replace each variable x1,x2,x3,x4 with some other variables u1,u2,u3,u4,

so that q(u1,u2,u3,u4) = ±(c1u1)^2 ± (c2u2)^2 ± (c3u3)^3 ± (c4u4)^2, so that the matrix for q

in the variables (u1,u2,u3,u4) is diag(±c1,±c2,±c3,±c4).

so if i have $2x_1x_4-6x_2x_3$
you said that when i have kab, then i should use use a' = a√k/2, b' = b√k/2
so for $2x_1x_4$ i use $x_1'=\frac{\sqrt{2}x1}{2}$ and
$x_4'=\frac{\sqrt{2}x4}{2}$
then i use
by multiplying x'1 and x'4 i dont get the proper form

**Deveno** · October 3rd 2011, 01:35 AM

suppose a' = a√k/2 and b' = b√k/2.

what is (a'+b')^2 - (a'-b')^2?

**transgalactic** · October 3rd 2011, 02:17 AM

$a'=\frac{{a\sqrt{k}}}{2} b'=\frac{{b\sqrt{k}}}{2}$

$(a'+b')^{2}-(a'-b')^{2}=(\frac{{a\sqrt{k}}}{2}+\frac{{b\sqrt{k}}}{ 2})^{2}-(\frac{{a\sqrt{k}}}{2}-\frac{{b\sqrt{k}}}{2})^{2}=2ab\sqrt{k}$

i understand

thanks

**Deveno** · October 3rd 2011, 02:54 AM

almost....

$(a'+b')^2 - (a'-b')^2 = (\frac{a\sqrt{k}}{2}+\frac{b\sqrt{k}}{2})^2 - (\frac{a\sqrt{k}}{2}-\frac{b\sqrt{k}}{2})^2$

$= (\frac{a^2k}{4} + \frac{2abk}{4} + \frac{b^2k}{4}) - (\frac{a^2k}{4} - \frac{2abk}{4} + \frac{b^2k}{4}) = kab$

**transgalactic** · October 3rd 2011, 04:30 AM

regarding part A)
in order to prove that its absolute positive we need to show that $T=T^*$ and that all of its eigen values
are positive

i cant see how can i get it from $T(T^{*}-3I)+2I=0$
or that its inner product

?

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