How to determine eigenvectors of symmetric circulant matrix {{A,B,B},{B,A,B},{B,B,A}}?












1












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I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .



I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .



But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.



From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:



$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$



Any help very appreciated.






linear-algebra eigenvalues-eigenvectors







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  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
    $endgroup$
    – amd
    Jan 16 at 23:06


















1












$begingroup$


I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .



I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .



But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.



From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:



$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$



Any help very appreciated.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$












  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
    $endgroup$
    – amd
    Jan 16 at 23:06
















1












1








1





$begingroup$


I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .



I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .



But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.



From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:



$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$



Any help very appreciated.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$




I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .



I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .



But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.



From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:



$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$



Any help very appreciated.






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






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asked Jan 16 at 20:06









fazanfazan

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  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
    $endgroup$
    – amd
    Jan 16 at 23:06




















  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
    $endgroup$
    – amd
    Jan 16 at 23:06


















$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06






$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06












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We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.



In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.



We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.






share|cite|improve this answer









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    $begingroup$

    We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.



    In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.



    We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.



      In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.



      We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.



        In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.



        We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.






        share|cite|improve this answer









        $endgroup$



        We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.



        In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.



        We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 20:47









        I like SerenaI like Serena

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