Eigenvector, solving Ax=b












0












$begingroup$


Hi I have another question where I don’t really know where to start....



Let A be a symmetric 3x3 matrix, with EV1 being (0,0,1) for $lambda_1$=0 and EV2 b=(2,1,0) for $lambda_2=1$. the third Eigenvalue be negativ.



With b from above give one solution for Ax=b



I honestly have no idea where to start, I can calculate the third eigenvector, and I know that the determinant of A is 0. I first thought of Cramer’s rule, but then I would have to divide by 0.



The only vague thing i could possibly think of is to go back to



$$Ax=lambda x $$



So i know lambda and x that would give me



A * (2,1,0)^t=1 * (2,10)



And this could only be true if A = I



So the solution would be x1=1 x2=1 and x3 = 0,



But i guess this is just stupid



I honestly have no idea...



Many thanks






linear-algebra







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You know that $b$ is an eigenvector of $A$ with eigenvalue $1$, correct? If so, then the equation $Ax=b$ has a solution in $x=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:39








  • 1




    $begingroup$
    Your problem is you have 2 Eigen values and the corresponding eigenvectors and you need to find the 3rd?
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:46






  • 2




    $begingroup$
    Well it follows from the definitions. We know that $b$ has eigenvalue $1$, so we have $Ab=1cdot b=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:47






  • 1




    $begingroup$
    Yes that's definitely a way.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:49






  • 1




    $begingroup$
    @Lily No, one eigenvalue of $A$ is $0$, so its determinant is also $0$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:57
















0












$begingroup$


Hi I have another question where I don’t really know where to start....



Let A be a symmetric 3x3 matrix, with EV1 being (0,0,1) for $lambda_1$=0 and EV2 b=(2,1,0) for $lambda_2=1$. the third Eigenvalue be negativ.



With b from above give one solution for Ax=b



I honestly have no idea where to start, I can calculate the third eigenvector, and I know that the determinant of A is 0. I first thought of Cramer’s rule, but then I would have to divide by 0.



The only vague thing i could possibly think of is to go back to



$$Ax=lambda x $$



So i know lambda and x that would give me



A * (2,1,0)^t=1 * (2,10)



And this could only be true if A = I



So the solution would be x1=1 x2=1 and x3 = 0,



But i guess this is just stupid



I honestly have no idea...



Many thanks






linear-algebra







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You know that $b$ is an eigenvector of $A$ with eigenvalue $1$, correct? If so, then the equation $Ax=b$ has a solution in $x=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:39








  • 1




    $begingroup$
    Your problem is you have 2 Eigen values and the corresponding eigenvectors and you need to find the 3rd?
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:46






  • 2




    $begingroup$
    Well it follows from the definitions. We know that $b$ has eigenvalue $1$, so we have $Ab=1cdot b=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:47






  • 1




    $begingroup$
    Yes that's definitely a way.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:49






  • 1




    $begingroup$
    @Lily No, one eigenvalue of $A$ is $0$, so its determinant is also $0$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:57














0












0








0





$begingroup$


Hi I have another question where I don’t really know where to start....



Let A be a symmetric 3x3 matrix, with EV1 being (0,0,1) for $lambda_1$=0 and EV2 b=(2,1,0) for $lambda_2=1$. the third Eigenvalue be negativ.



With b from above give one solution for Ax=b



I honestly have no idea where to start, I can calculate the third eigenvector, and I know that the determinant of A is 0. I first thought of Cramer’s rule, but then I would have to divide by 0.



The only vague thing i could possibly think of is to go back to



$$Ax=lambda x $$



So i know lambda and x that would give me



A * (2,1,0)^t=1 * (2,10)



And this could only be true if A = I



So the solution would be x1=1 x2=1 and x3 = 0,



But i guess this is just stupid



I honestly have no idea...



Many thanks






linear-algebra







share|cite|improve this question









$endgroup$




Hi I have another question where I don’t really know where to start....



Let A be a symmetric 3x3 matrix, with EV1 being (0,0,1) for $lambda_1$=0 and EV2 b=(2,1,0) for $lambda_2=1$. the third Eigenvalue be negativ.



With b from above give one solution for Ax=b



I honestly have no idea where to start, I can calculate the third eigenvector, and I know that the determinant of A is 0. I first thought of Cramer’s rule, but then I would have to divide by 0.



The only vague thing i could possibly think of is to go back to



$$Ax=lambda x $$



So i know lambda and x that would give me



A * (2,1,0)^t=1 * (2,10)



And this could only be true if A = I



So the solution would be x1=1 x2=1 and x3 = 0,



But i guess this is just stupid



I honestly have no idea...



Many thanks






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 16:28









LillysLillys

778




778








  • 1




    $begingroup$
    You know that $b$ is an eigenvector of $A$ with eigenvalue $1$, correct? If so, then the equation $Ax=b$ has a solution in $x=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:39








  • 1




    $begingroup$
    Your problem is you have 2 Eigen values and the corresponding eigenvectors and you need to find the 3rd?
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:46






  • 2




    $begingroup$
    Well it follows from the definitions. We know that $b$ has eigenvalue $1$, so we have $Ab=1cdot b=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:47






  • 1




    $begingroup$
    Yes that's definitely a way.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:49






  • 1




    $begingroup$
    @Lily No, one eigenvalue of $A$ is $0$, so its determinant is also $0$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:57














  • 1




    $begingroup$
    You know that $b$ is an eigenvector of $A$ with eigenvalue $1$, correct? If so, then the equation $Ax=b$ has a solution in $x=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:39








  • 1




    $begingroup$
    Your problem is you have 2 Eigen values and the corresponding eigenvectors and you need to find the 3rd?
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:46






  • 2




    $begingroup$
    Well it follows from the definitions. We know that $b$ has eigenvalue $1$, so we have $Ab=1cdot b=b$.
    $endgroup$
    – Dave
    Dec 11 '18 at 16:47






  • 1




    $begingroup$
    Yes that's definitely a way.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 16:49






  • 1




    $begingroup$
    @Lily No, one eigenvalue of $A$ is $0$, so its determinant is also $0$
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 16:57








1




1




$begingroup$
You know that $b$ is an eigenvector of $A$ with eigenvalue $1$, correct? If so, then the equation $Ax=b$ has a solution in $x=b$.
$endgroup$
– Dave
Dec 11 '18 at 16:39






$begingroup$
You know that $b$ is an eigenvector of $A$ with eigenvalue $1$, correct? If so, then the equation $Ax=b$ has a solution in $x=b$.
$endgroup$
– Dave
Dec 11 '18 at 16:39






1




1




$begingroup$
Your problem is you have 2 Eigen values and the corresponding eigenvectors and you need to find the 3rd?
$endgroup$
– mm-crj
Dec 11 '18 at 16:46




$begingroup$
Your problem is you have 2 Eigen values and the corresponding eigenvectors and you need to find the 3rd?
$endgroup$
– mm-crj
Dec 11 '18 at 16:46




2




2




$begingroup$
Well it follows from the definitions. We know that $b$ has eigenvalue $1$, so we have $Ab=1cdot b=b$.
$endgroup$
– Dave
Dec 11 '18 at 16:47




$begingroup$
Well it follows from the definitions. We know that $b$ has eigenvalue $1$, so we have $Ab=1cdot b=b$.
$endgroup$
– Dave
Dec 11 '18 at 16:47




1




1




$begingroup$
Yes that's definitely a way.
$endgroup$
– mm-crj
Dec 11 '18 at 16:49




$begingroup$
Yes that's definitely a way.
$endgroup$
– mm-crj
Dec 11 '18 at 16:49




1




1




$begingroup$
@Lily No, one eigenvalue of $A$ is $0$, so its determinant is also $0$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:57




$begingroup$
@Lily No, one eigenvalue of $A$ is $0$, so its determinant is also $0$
$endgroup$
– Shubham Johri
Dec 11 '18 at 16:57










1 Answer
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$begingroup$

For the eigenvector $vec b, Avec b=lambdavec bimpliesvec b=Acdotfrac{vec b}lambda (lambdane0)$



So one solution for the system $Avec x=vec b$ for eigenvector $vec b=[2,1,0]^T$ corresponding to the non-zero eigenvalue $lambda=1$ is $vec b/lambda$. Additionally, since $det A=0$, the system $Avec x=vec b$ can have either no or infinitely many solutions. Since one solution $vec b/lambda$ has been shown to exist, we know that infinitely many solutions exist.



Note that $vec x=frac{vec b}lambda+kcdot[0,0,1]^T,kinBbb R$, will always be a solution to the system as $A(frac{vec b}lambda+kcdot[0,0,1]^T)=vec b+kcdot A[0,0,1]^T=vec b because A[0,0,1]^T=0$






share|cite|improve this answer











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    $begingroup$

    For the eigenvector $vec b, Avec b=lambdavec bimpliesvec b=Acdotfrac{vec b}lambda (lambdane0)$



    So one solution for the system $Avec x=vec b$ for eigenvector $vec b=[2,1,0]^T$ corresponding to the non-zero eigenvalue $lambda=1$ is $vec b/lambda$. Additionally, since $det A=0$, the system $Avec x=vec b$ can have either no or infinitely many solutions. Since one solution $vec b/lambda$ has been shown to exist, we know that infinitely many solutions exist.



    Note that $vec x=frac{vec b}lambda+kcdot[0,0,1]^T,kinBbb R$, will always be a solution to the system as $A(frac{vec b}lambda+kcdot[0,0,1]^T)=vec b+kcdot A[0,0,1]^T=vec b because A[0,0,1]^T=0$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      For the eigenvector $vec b, Avec b=lambdavec bimpliesvec b=Acdotfrac{vec b}lambda (lambdane0)$



      So one solution for the system $Avec x=vec b$ for eigenvector $vec b=[2,1,0]^T$ corresponding to the non-zero eigenvalue $lambda=1$ is $vec b/lambda$. Additionally, since $det A=0$, the system $Avec x=vec b$ can have either no or infinitely many solutions. Since one solution $vec b/lambda$ has been shown to exist, we know that infinitely many solutions exist.



      Note that $vec x=frac{vec b}lambda+kcdot[0,0,1]^T,kinBbb R$, will always be a solution to the system as $A(frac{vec b}lambda+kcdot[0,0,1]^T)=vec b+kcdot A[0,0,1]^T=vec b because A[0,0,1]^T=0$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        For the eigenvector $vec b, Avec b=lambdavec bimpliesvec b=Acdotfrac{vec b}lambda (lambdane0)$



        So one solution for the system $Avec x=vec b$ for eigenvector $vec b=[2,1,0]^T$ corresponding to the non-zero eigenvalue $lambda=1$ is $vec b/lambda$. Additionally, since $det A=0$, the system $Avec x=vec b$ can have either no or infinitely many solutions. Since one solution $vec b/lambda$ has been shown to exist, we know that infinitely many solutions exist.



        Note that $vec x=frac{vec b}lambda+kcdot[0,0,1]^T,kinBbb R$, will always be a solution to the system as $A(frac{vec b}lambda+kcdot[0,0,1]^T)=vec b+kcdot A[0,0,1]^T=vec b because A[0,0,1]^T=0$






        share|cite|improve this answer











        $endgroup$



        For the eigenvector $vec b, Avec b=lambdavec bimpliesvec b=Acdotfrac{vec b}lambda (lambdane0)$



        So one solution for the system $Avec x=vec b$ for eigenvector $vec b=[2,1,0]^T$ corresponding to the non-zero eigenvalue $lambda=1$ is $vec b/lambda$. Additionally, since $det A=0$, the system $Avec x=vec b$ can have either no or infinitely many solutions. Since one solution $vec b/lambda$ has been shown to exist, we know that infinitely many solutions exist.



        Note that $vec x=frac{vec b}lambda+kcdot[0,0,1]^T,kinBbb R$, will always be a solution to the system as $A(frac{vec b}lambda+kcdot[0,0,1]^T)=vec b+kcdot A[0,0,1]^T=vec b because A[0,0,1]^T=0$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 17:09

























        answered Dec 11 '18 at 17:04









        Shubham JohriShubham Johri

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