What is the solution for these two equalities $ x = lambda(Ax) $ and $ x^TAx=1 $?
$begingroup$
Let $A in mathbb{R}^{4 times 4}$ be a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.
What is the solution for $x$ when $lambda >0$
$$ x = lambda(Ax) $$ and $$ x^TAx=1 $$
linear-algebra matrices matrix-calculus symmetric-matrices
asked Dec 17 '18 at 5:06
SepideSepide
2888
$endgroup$
add a comment |
$begingroup$
Let $A in mathbb{R}^{4 times 4}$ be a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.
What is the solution for $x$ when $lambda >0$
$$ x = lambda(Ax) $$ and $$ x^TAx=1 $$
linear-algebra matrices matrix-calculus symmetric-matrices
asked Dec 17 '18 at 5:06
SepideSepide
2888
$endgroup$
add a comment |
$begingroup$
Let $A in mathbb{R}^{4 times 4}$ be a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.
What is the solution for $x$ when $lambda >0$
$$ x = lambda(Ax) $$ and $$ x^TAx=1 $$
linear-algebra matrices matrix-calculus symmetric-matrices
asked Dec 17 '18 at 5:06
SepideSepide
2888
$endgroup$
Let $A in mathbb{R}^{4 times 4}$ be a symmetric matrix with two distinct positive eigenvalues and other eigenvalues of $A$ are nonpositive.
What is the solution for $x$ when $lambda >0$
$$ x = lambda(Ax) $$ and $$ x^TAx=1 $$
linear-algebra matrices matrix-calculus symmetric-matrices
linear-algebra matrices matrix-calculus symmetric-matrices
asked Dec 17 '18 at 5:06
SepideSepide
2888
asked Dec 17 '18 at 5:06
SepideSepide
2888
asked Dec 17 '18 at 5:06
SepideSepide
2888
asked Dec 17 '18 at 5:06
SepideSepide
2888
asked Dec 17 '18 at 5:06
SepideSepide
2888
2888
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose that $x=lambda Ax$, then
$$lambda Ax-x=0$$
$$lambda(Ax-lambda^{-1}x)=0$$
$$(A-lambda^{-1}I)x=0$$
So $x$ is an eigenvector with eigenvalue $lambda^{-1}>0$. Moreover,
$$x^T Ax=lambda^{-1}x^T x=1 Rightarrow x^T x=||x||^2=lambda$$
So any solutions, $x$, must be eigenvectors with (positive) eigenvalue $lambda^{-1}$ and norm $||x||=sqrt{lambda}$. If the unit eigenvectors of $A$ are $v_1$ and $v_2$
with the corresponding positve eigenvalues $mu_1$ and $mu_2$, then the solutions to your system of equations are:
$$(x,lambda) in {(sqrt{mu_1^{-1}}v_1,mu_1^{-1}),(sqrt{mu_2^{-1}}v_1,mu_2^{-1}) }$$
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
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1 Answer
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1 Answer
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$begingroup$
Suppose that $x=lambda Ax$, then
$$lambda Ax-x=0$$
$$lambda(Ax-lambda^{-1}x)=0$$
$$(A-lambda^{-1}I)x=0$$
So $x$ is an eigenvector with eigenvalue $lambda^{-1}>0$. Moreover,
$$x^T Ax=lambda^{-1}x^T x=1 Rightarrow x^T x=||x||^2=lambda$$
So any solutions, $x$, must be eigenvectors with (positive) eigenvalue $lambda^{-1}$ and norm $||x||=sqrt{lambda}$. If the unit eigenvectors of $A$ are $v_1$ and $v_2$
with the corresponding positve eigenvalues $mu_1$ and $mu_2$, then the solutions to your system of equations are:
$$(x,lambda) in {(sqrt{mu_1^{-1}}v_1,mu_1^{-1}),(sqrt{mu_2^{-1}}v_1,mu_2^{-1}) }$$
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
$endgroup$
add a comment |
$begingroup$
Suppose that $x=lambda Ax$, then
$$lambda Ax-x=0$$
$$lambda(Ax-lambda^{-1}x)=0$$
$$(A-lambda^{-1}I)x=0$$
So $x$ is an eigenvector with eigenvalue $lambda^{-1}>0$. Moreover,
$$x^T Ax=lambda^{-1}x^T x=1 Rightarrow x^T x=||x||^2=lambda$$
So any solutions, $x$, must be eigenvectors with (positive) eigenvalue $lambda^{-1}$ and norm $||x||=sqrt{lambda}$. If the unit eigenvectors of $A$ are $v_1$ and $v_2$
with the corresponding positve eigenvalues $mu_1$ and $mu_2$, then the solutions to your system of equations are:
$$(x,lambda) in {(sqrt{mu_1^{-1}}v_1,mu_1^{-1}),(sqrt{mu_2^{-1}}v_1,mu_2^{-1}) }$$
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
$endgroup$
add a comment |
$begingroup$
Suppose that $x=lambda Ax$, then
$$lambda Ax-x=0$$
$$lambda(Ax-lambda^{-1}x)=0$$
$$(A-lambda^{-1}I)x=0$$
So $x$ is an eigenvector with eigenvalue $lambda^{-1}>0$. Moreover,
$$x^T Ax=lambda^{-1}x^T x=1 Rightarrow x^T x=||x||^2=lambda$$
So any solutions, $x$, must be eigenvectors with (positive) eigenvalue $lambda^{-1}$ and norm $||x||=sqrt{lambda}$. If the unit eigenvectors of $A$ are $v_1$ and $v_2$
with the corresponding positve eigenvalues $mu_1$ and $mu_2$, then the solutions to your system of equations are:
$$(x,lambda) in {(sqrt{mu_1^{-1}}v_1,mu_1^{-1}),(sqrt{mu_2^{-1}}v_1,mu_2^{-1}) }$$
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
$endgroup$
Suppose that $x=lambda Ax$, then
$$lambda Ax-x=0$$
$$lambda(Ax-lambda^{-1}x)=0$$
$$(A-lambda^{-1}I)x=0$$
So $x$ is an eigenvector with eigenvalue $lambda^{-1}>0$. Moreover,
$$x^T Ax=lambda^{-1}x^T x=1 Rightarrow x^T x=||x||^2=lambda$$
So any solutions, $x$, must be eigenvectors with (positive) eigenvalue $lambda^{-1}$ and norm $||x||=sqrt{lambda}$. If the unit eigenvectors of $A$ are $v_1$ and $v_2$
with the corresponding positve eigenvalues $mu_1$ and $mu_2$, then the solutions to your system of equations are:
$$(x,lambda) in {(sqrt{mu_1^{-1}}v_1,mu_1^{-1}),(sqrt{mu_2^{-1}}v_1,mu_2^{-1}) }$$
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
answered Dec 17 '18 at 9:47
clmundergradclmundergrad
1665
1665
add a comment |
add a comment |
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