How do I find a vector x such that $x'Ax < 0 $ or make the matrix $A $ no longer SPD?
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I'm given a matrix a 3x3 matrix $A $ , where I need to find a vector x such that $x'Ax $ it gives me a negative number (or x'Ax < 0). I was thinking I have a vector x set to x = [a, b, c] and perform gaussian elimination but not sure where to go from there.
Any tips on how to approach this would be great!
numerical-methods numerical-linear-algebra
edited Nov 14 at 10:56
dmtri
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asked Nov 14 at 10:44
user208691
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I'm given a matrix a 3x3 matrix $A $ , where I need to find a vector x such that $x'Ax $ it gives me a negative number (or x'Ax < 0). I was thinking I have a vector x set to x = [a, b, c] and perform gaussian elimination but not sure where to go from there.
Any tips on how to approach this would be great!
numerical-methods numerical-linear-algebra
edited Nov 14 at 10:56
dmtri
1,2261521
asked Nov 14 at 10:44
user208691
6
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm given a matrix a 3x3 matrix $A $ , where I need to find a vector x such that $x'Ax $ it gives me a negative number (or x'Ax < 0). I was thinking I have a vector x set to x = [a, b, c] and perform gaussian elimination but not sure where to go from there.
Any tips on how to approach this would be great!
numerical-methods numerical-linear-algebra
edited Nov 14 at 10:56
dmtri
1,2261521
asked Nov 14 at 10:44
user208691
6
I'm given a matrix a 3x3 matrix $A $ , where I need to find a vector x such that $x'Ax $ it gives me a negative number (or x'Ax < 0). I was thinking I have a vector x set to x = [a, b, c] and perform gaussian elimination but not sure where to go from there.
Any tips on how to approach this would be great!
numerical-methods numerical-linear-algebra
numerical-methods numerical-linear-algebra
edited Nov 14 at 10:56
dmtri
1,2261521
asked Nov 14 at 10:44
user208691
6
edited Nov 14 at 10:56
dmtri
1,2261521
asked Nov 14 at 10:44
user208691
6
edited Nov 14 at 10:56
dmtri
1,2261521
edited Nov 14 at 10:56
dmtri
1,2261521
edited Nov 14 at 10:56
dmtri
1,2261521
1,2261521
asked Nov 14 at 10:44
user208691
6
asked Nov 14 at 10:44
user208691
6
asked Nov 14 at 10:44
user208691
6
6
add a comment |
add a comment |
1 Answer
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First of all, find the eigenvalues $lambda_k$ and eigenvectors $V_k$ associated with them. Two cases :
(a) either all $lambda_k$s are $geq 0$ , in such a case, you will never find an $X$ such that $X^TAX <0$.
(b) or one at least of the eigenvalues, say $lambda_1$ is $<0$, in this case, take $X=V_1$. Indeed, $X^TAX=V_1^T(AV_1)=V_1^T(lambda_1 V_1)$ (by definition of an eigenvector), which gives $lambda_1 |V_1|^2 < 0$. That's all.
Considering the alternative you give in your title (find a way to make $A$ "no longer Symmetric Positive Definite"), you just have to "shift left" the spectrum of $A$. For example, if Spec$(A)={2,3,7}$, replace $A$ by, say, $A-2.001 I_3$, the spectrum of the new matrix will be shifted by $-2$ , i.e., will become${-0.001,0.999,4.999}$, and we are now in case (b)
answered Nov 17 at 22:18
Jean Marie
28.2k41848
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
First of all, find the eigenvalues $lambda_k$ and eigenvectors $V_k$ associated with them. Two cases :
(a) either all $lambda_k$s are $geq 0$ , in such a case, you will never find an $X$ such that $X^TAX <0$.
(b) or one at least of the eigenvalues, say $lambda_1$ is $<0$, in this case, take $X=V_1$. Indeed, $X^TAX=V_1^T(AV_1)=V_1^T(lambda_1 V_1)$ (by definition of an eigenvector), which gives $lambda_1 |V_1|^2 < 0$. That's all.
Considering the alternative you give in your title (find a way to make $A$ "no longer Symmetric Positive Definite"), you just have to "shift left" the spectrum of $A$. For example, if Spec$(A)={2,3,7}$, replace $A$ by, say, $A-2.001 I_3$, the spectrum of the new matrix will be shifted by $-2$ , i.e., will become${-0.001,0.999,4.999}$, and we are now in case (b)
answered Nov 17 at 22:18
Jean Marie
28.2k41848
add a comment |
up vote
0
down vote
First of all, find the eigenvalues $lambda_k$ and eigenvectors $V_k$ associated with them. Two cases :
(a) either all $lambda_k$s are $geq 0$ , in such a case, you will never find an $X$ such that $X^TAX <0$.
(b) or one at least of the eigenvalues, say $lambda_1$ is $<0$, in this case, take $X=V_1$. Indeed, $X^TAX=V_1^T(AV_1)=V_1^T(lambda_1 V_1)$ (by definition of an eigenvector), which gives $lambda_1 |V_1|^2 < 0$. That's all.
Considering the alternative you give in your title (find a way to make $A$ "no longer Symmetric Positive Definite"), you just have to "shift left" the spectrum of $A$. For example, if Spec$(A)={2,3,7}$, replace $A$ by, say, $A-2.001 I_3$, the spectrum of the new matrix will be shifted by $-2$ , i.e., will become${-0.001,0.999,4.999}$, and we are now in case (b)
answered Nov 17 at 22:18
Jean Marie
28.2k41848
add a comment |
up vote
0
down vote
up vote
0
down vote
First of all, find the eigenvalues $lambda_k$ and eigenvectors $V_k$ associated with them. Two cases :
(a) either all $lambda_k$s are $geq 0$ , in such a case, you will never find an $X$ such that $X^TAX <0$.
(b) or one at least of the eigenvalues, say $lambda_1$ is $<0$, in this case, take $X=V_1$. Indeed, $X^TAX=V_1^T(AV_1)=V_1^T(lambda_1 V_1)$ (by definition of an eigenvector), which gives $lambda_1 |V_1|^2 < 0$. That's all.
Considering the alternative you give in your title (find a way to make $A$ "no longer Symmetric Positive Definite"), you just have to "shift left" the spectrum of $A$. For example, if Spec$(A)={2,3,7}$, replace $A$ by, say, $A-2.001 I_3$, the spectrum of the new matrix will be shifted by $-2$ , i.e., will become${-0.001,0.999,4.999}$, and we are now in case (b)
answered Nov 17 at 22:18
Jean Marie
28.2k41848
First of all, find the eigenvalues $lambda_k$ and eigenvectors $V_k$ associated with them. Two cases :
(a) either all $lambda_k$s are $geq 0$ , in such a case, you will never find an $X$ such that $X^TAX <0$.
(b) or one at least of the eigenvalues, say $lambda_1$ is $<0$, in this case, take $X=V_1$. Indeed, $X^TAX=V_1^T(AV_1)=V_1^T(lambda_1 V_1)$ (by definition of an eigenvector), which gives $lambda_1 |V_1|^2 < 0$. That's all.
Considering the alternative you give in your title (find a way to make $A$ "no longer Symmetric Positive Definite"), you just have to "shift left" the spectrum of $A$. For example, if Spec$(A)={2,3,7}$, replace $A$ by, say, $A-2.001 I_3$, the spectrum of the new matrix will be shifted by $-2$ , i.e., will become${-0.001,0.999,4.999}$, and we are now in case (b)
answered Nov 17 at 22:18
Jean Marie
28.2k41848
edited Nov 18 at 7:04
answered Nov 17 at 22:18
Jean Marie
28.2k41848
answered Nov 17 at 22:18
Jean Marie
28.2k41848
answered Nov 17 at 22:18
Jean Marie
28.2k41848
28.2k41848
add a comment |
add a comment |
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