Min and max eigenvalues of a quadratic form
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I have the following quadratic form :
$q(x,y,z)=x(y+4z)+y(x-2z)+x^2$, represented by the symmetric matrix $begin{bmatrix}1&1&2\1&0&-1\2&-1&0end{bmatrix}$.
I am asked to find its signature, which appears to be $(2,1)$. And then, I am asked to prove that all of its matrix's eigenvalues are in $]-3,3[$ without computing them. I can't see how to do this.
eigenvalues-eigenvectors quadratic-forms
asked Nov 14 at 10:30
James Well
511410
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I have the following quadratic form :
$q(x,y,z)=x(y+4z)+y(x-2z)+x^2$, represented by the symmetric matrix $begin{bmatrix}1&1&2\1&0&-1\2&-1&0end{bmatrix}$.
I am asked to find its signature, which appears to be $(2,1)$. And then, I am asked to prove that all of its matrix's eigenvalues are in $]-3,3[$ without computing them. I can't see how to do this.
eigenvalues-eigenvectors quadratic-forms
asked Nov 14 at 10:30
James Well
511410
"...appears to be"? Can't you just calculate it? And what does "computing them" (the eigenvalues) mean? Without finding them explicitly but with the char. pol. it looks like a more or less easy exercise in Calculus I, IVT and etc.
– DonAntonio
Nov 14 at 10:35
Use Gershgorin circle theorem
– Yadati Kiran
Nov 14 at 10:46
@YadatiKiran Gershgorin's Theorem doesn't give so tight a bound for the eigenvalues of the matrix. As far as I can see it only gives $;(-3,4);$...
– DonAntonio
Nov 14 at 10:49
The bound is $(-2,4)$. We get three discs $D(1,3), D(0,2)$ and $D(0,3)$
– Yadati Kiran
Nov 14 at 11:00
We can also say $sum|lambda-a_{ii}|leqsum R_i implies |lambda-1|+|lambda|+|lambda|leq8 implies dfrac{-7}{3}leqlambdaleq3$
– Yadati Kiran
Nov 14 at 11:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following quadratic form :
$q(x,y,z)=x(y+4z)+y(x-2z)+x^2$, represented by the symmetric matrix $begin{bmatrix}1&1&2\1&0&-1\2&-1&0end{bmatrix}$.
I am asked to find its signature, which appears to be $(2,1)$. And then, I am asked to prove that all of its matrix's eigenvalues are in $]-3,3[$ without computing them. I can't see how to do this.
eigenvalues-eigenvectors quadratic-forms
asked Nov 14 at 10:30
James Well
511410
I have the following quadratic form :
$q(x,y,z)=x(y+4z)+y(x-2z)+x^2$, represented by the symmetric matrix $begin{bmatrix}1&1&2\1&0&-1\2&-1&0end{bmatrix}$.
I am asked to find its signature, which appears to be $(2,1)$. And then, I am asked to prove that all of its matrix's eigenvalues are in $]-3,3[$ without computing them. I can't see how to do this.
eigenvalues-eigenvectors quadratic-forms
eigenvalues-eigenvectors quadratic-forms
asked Nov 14 at 10:30
James Well
511410
asked Nov 14 at 10:30
James Well
511410
asked Nov 14 at 10:30
James Well
511410
asked Nov 14 at 10:30
James Well
511410
asked Nov 14 at 10:30
James Well
511410
511410
"...appears to be"? Can't you just calculate it? And what does "computing them" (the eigenvalues) mean? Without finding them explicitly but with the char. pol. it looks like a more or less easy exercise in Calculus I, IVT and etc.
– DonAntonio
Nov 14 at 10:35
Use Gershgorin circle theorem
– Yadati Kiran
Nov 14 at 10:46
@YadatiKiran Gershgorin's Theorem doesn't give so tight a bound for the eigenvalues of the matrix. As far as I can see it only gives $;(-3,4);$...
– DonAntonio
Nov 14 at 10:49
The bound is $(-2,4)$. We get three discs $D(1,3), D(0,2)$ and $D(0,3)$
– Yadati Kiran
Nov 14 at 11:00
We can also say $sum|lambda-a_{ii}|leqsum R_i implies |lambda-1|+|lambda|+|lambda|leq8 implies dfrac{-7}{3}leqlambdaleq3$
– Yadati Kiran
Nov 14 at 11:07
add a comment |
"...appears to be"? Can't you just calculate it? And what does "computing them" (the eigenvalues) mean? Without finding them explicitly but with the char. pol. it looks like a more or less easy exercise in Calculus I, IVT and etc.
– DonAntonio
Nov 14 at 10:35
Use Gershgorin circle theorem
– Yadati Kiran
Nov 14 at 10:46
@YadatiKiran Gershgorin's Theorem doesn't give so tight a bound for the eigenvalues of the matrix. As far as I can see it only gives $;(-3,4);$...
– DonAntonio
Nov 14 at 10:49
The bound is $(-2,4)$. We get three discs $D(1,3), D(0,2)$ and $D(0,3)$
– Yadati Kiran
Nov 14 at 11:00
We can also say $sum|lambda-a_{ii}|leqsum R_i implies |lambda-1|+|lambda|+|lambda|leq8 implies dfrac{-7}{3}leqlambdaleq3$
– Yadati Kiran
Nov 14 at 11:07
"...appears to be"? Can't you just calculate it? And what does "computing them" (the eigenvalues) mean? Without finding them explicitly but with the char. pol. it looks like a more or less easy exercise in Calculus I, IVT and etc.
– DonAntonio
Nov 14 at 10:35
"...appears to be"? Can't you just calculate it? And what does "computing them" (the eigenvalues) mean? Without finding them explicitly but with the char. pol. it looks like a more or less easy exercise in Calculus I, IVT and etc.
– DonAntonio
Nov 14 at 10:35
Use Gershgorin circle theorem
– Yadati Kiran
Nov 14 at 10:46
Use Gershgorin circle theorem
– Yadati Kiran
Nov 14 at 10:46
@YadatiKiran Gershgorin's Theorem doesn't give so tight a bound for the eigenvalues of the matrix. As far as I can see it only gives $;(-3,4);$...
– DonAntonio
Nov 14 at 10:49
@YadatiKiran Gershgorin's Theorem doesn't give so tight a bound for the eigenvalues of the matrix. As far as I can see it only gives $;(-3,4);$...
– DonAntonio
Nov 14 at 10:49
The bound is $(-2,4)$. We get three discs $D(1,3), D(0,2)$ and $D(0,3)$
– Yadati Kiran
Nov 14 at 11:00
The bound is $(-2,4)$. We get three discs $D(1,3), D(0,2)$ and $D(0,3)$
– Yadati Kiran
Nov 14 at 11:00
We can also say $sum|lambda-a_{ii}|leqsum R_i implies |lambda-1|+|lambda|+|lambda|leq8 implies dfrac{-7}{3}leqlambdaleq3$
– Yadati Kiran
Nov 14 at 11:07
We can also say $sum|lambda-a_{ii}|leqsum R_i implies |lambda-1|+|lambda|+|lambda|leq8 implies dfrac{-7}{3}leqlambdaleq3$
– Yadati Kiran
Nov 14 at 11:07
add a comment |
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"...appears to be"? Can't you just calculate it? And what does "computing them" (the eigenvalues) mean? Without finding them explicitly but with the char. pol. it looks like a more or less easy exercise in Calculus I, IVT and etc.
– DonAntonio
Nov 14 at 10:35
Use Gershgorin circle theorem
– Yadati Kiran
Nov 14 at 10:46
@YadatiKiran Gershgorin's Theorem doesn't give so tight a bound for the eigenvalues of the matrix. As far as I can see it only gives $;(-3,4);$...
– DonAntonio
Nov 14 at 10:49
The bound is $(-2,4)$. We get three discs $D(1,3), D(0,2)$ and $D(0,3)$
– Yadati Kiran
Nov 14 at 11:00
We can also say $sum|lambda-a_{ii}|leqsum R_i implies |lambda-1|+|lambda|+|lambda|leq8 implies dfrac{-7}{3}leqlambdaleq3$
– Yadati Kiran
Nov 14 at 11:07