How to deduce third eigenvalue/eigenvector pair of a matrix from first two pairs and determinant












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Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?






linear-algebra matrices







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  • 1




    $begingroup$
    The product of the three eigenvalues is the determinant.
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    – T. Bongers
    Dec 11 '18 at 2:25












  • $begingroup$
    I know that, but how do you find its corresponding eigenvector?
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:26










  • $begingroup$
    Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
    $endgroup$
    – Morgan Rodgers
    Dec 11 '18 at 2:28












  • $begingroup$
    You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:31










  • $begingroup$
    There’s no way to determine the missing eigenspace without more information about $A$.
    $endgroup$
    – amd
    Dec 11 '18 at 2:50


















-1












$begingroup$


Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The product of the three eigenvalues is the determinant.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 2:25












  • $begingroup$
    I know that, but how do you find its corresponding eigenvector?
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:26










  • $begingroup$
    Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
    $endgroup$
    – Morgan Rodgers
    Dec 11 '18 at 2:28












  • $begingroup$
    You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:31










  • $begingroup$
    There’s no way to determine the missing eigenspace without more information about $A$.
    $endgroup$
    – amd
    Dec 11 '18 at 2:50
















-1












-1








-1





$begingroup$


Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?






linear-algebra matrices







share|cite|improve this question











$endgroup$




Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 2:31







BuluBestTapu

















asked Dec 11 '18 at 2:25









BuluBestTapuBuluBestTapu

778




778








  • 1




    $begingroup$
    The product of the three eigenvalues is the determinant.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 2:25












  • $begingroup$
    I know that, but how do you find its corresponding eigenvector?
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:26










  • $begingroup$
    Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
    $endgroup$
    – Morgan Rodgers
    Dec 11 '18 at 2:28












  • $begingroup$
    You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:31










  • $begingroup$
    There’s no way to determine the missing eigenspace without more information about $A$.
    $endgroup$
    – amd
    Dec 11 '18 at 2:50
















  • 1




    $begingroup$
    The product of the three eigenvalues is the determinant.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 2:25












  • $begingroup$
    I know that, but how do you find its corresponding eigenvector?
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:26










  • $begingroup$
    Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
    $endgroup$
    – Morgan Rodgers
    Dec 11 '18 at 2:28












  • $begingroup$
    You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
    $endgroup$
    – BuluBestTapu
    Dec 11 '18 at 2:31










  • $begingroup$
    There’s no way to determine the missing eigenspace without more information about $A$.
    $endgroup$
    – amd
    Dec 11 '18 at 2:50










1




1




$begingroup$
The product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25






$begingroup$
The product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25














$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26




$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26












$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28






$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28














$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31




$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31












$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50






$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50












1 Answer
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$begingroup$

Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.



So, you miss 2 relations.



EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".



Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.



Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".






share|cite|improve this answer











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    $begingroup$

    Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.



    So, you miss 2 relations.



    EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".



    Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.



    Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.



      So, you miss 2 relations.



      EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".



      Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.



      Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.



        So, you miss 2 relations.



        EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".



        Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.



        Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".






        share|cite|improve this answer











        $endgroup$



        Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.



        So, you miss 2 relations.



        EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".



        Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.



        Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 11:17

























        answered Dec 11 '18 at 10:41









        loup blancloup blanc

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        23.4k21851






























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