How to find Eigenvalue with Characteristic Equation?












0












$begingroup$


When my book explains using the characteristic equation to find eigenvalues, it gives this example.



Find the eigenvalues and eigenvectors of

A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$



= $(lambda - 2)^3$



It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)






linear-algebra







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$endgroup$












  • $begingroup$
    Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
    $endgroup$
    – Federico
    Dec 4 '18 at 19:56










  • $begingroup$
    fixed, thank you
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:58
















0












$begingroup$


When my book explains using the characteristic equation to find eigenvalues, it gives this example.



Find the eigenvalues and eigenvectors of

A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$



= $(lambda - 2)^3$



It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
    $endgroup$
    – Federico
    Dec 4 '18 at 19:56










  • $begingroup$
    fixed, thank you
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:58














0












0








0





$begingroup$


When my book explains using the characteristic equation to find eigenvalues, it gives this example.



Find the eigenvalues and eigenvectors of

A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$



= $(lambda - 2)^3$



It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)






linear-algebra







share|cite|improve this question











$endgroup$




When my book explains using the characteristic equation to find eigenvalues, it gives this example.



Find the eigenvalues and eigenvectors of

A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$



= $(lambda - 2)^3$



It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 19:58







Evan Kim

















asked Dec 4 '18 at 19:37









Evan KimEvan Kim

998




998












  • $begingroup$
    Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
    $endgroup$
    – Federico
    Dec 4 '18 at 19:56










  • $begingroup$
    fixed, thank you
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:58


















  • $begingroup$
    Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
    $endgroup$
    – Federico
    Dec 4 '18 at 19:56










  • $begingroup$
    fixed, thank you
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:58
















$begingroup$
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
$endgroup$
– Federico
Dec 4 '18 at 19:56




$begingroup$
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
$endgroup$
– Federico
Dec 4 '18 at 19:56












$begingroup$
fixed, thank you
$endgroup$
– Evan Kim
Dec 4 '18 at 19:58




$begingroup$
fixed, thank you
$endgroup$
– Evan Kim
Dec 4 '18 at 19:58










3 Answers
3






active

oldest

votes


















2












$begingroup$

The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.



Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:52










  • $begingroup$
    You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
    $endgroup$
    – Federico
    Dec 4 '18 at 19:53










  • $begingroup$
    @Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
    $endgroup$
    – user376343
    Dec 4 '18 at 19:54



















1












$begingroup$

A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :



$$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$



If you miss on that, calculate it the explicit way, by :



$$|lambda I - A| =
begin{vmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$



$$Rightarrow$$



$$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$



The eigenvalues then, would simply be :



$$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:59










  • $begingroup$
    @EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
    $endgroup$
    – Rebellos
    Dec 4 '18 at 20:03





















0












$begingroup$

In fact, your matrix is special. it makes the computation easier.



Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
$$(x,y,z)ne (0,0,0)$$
and
$$2x+y=kx$$
$$2y=ky$$
$$2z=kz$$
but
$$kne 2implies z=y=x=0$$



which is in contradiction with the assumption.






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.



    Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:52










    • $begingroup$
      You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
      $endgroup$
      – Federico
      Dec 4 '18 at 19:53










    • $begingroup$
      @Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
      $endgroup$
      – user376343
      Dec 4 '18 at 19:54
















    2












    $begingroup$

    The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.



    Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:52










    • $begingroup$
      You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
      $endgroup$
      – Federico
      Dec 4 '18 at 19:53










    • $begingroup$
      @Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
      $endgroup$
      – user376343
      Dec 4 '18 at 19:54














    2












    2








    2





    $begingroup$

    The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.



    Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.






    share|cite|improve this answer











    $endgroup$



    The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.



    Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 4 '18 at 19:54

























    answered Dec 4 '18 at 19:41









    FedericoFederico

    4,984514




    4,984514








    • 1




      $begingroup$
      So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:52










    • $begingroup$
      You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
      $endgroup$
      – Federico
      Dec 4 '18 at 19:53










    • $begingroup$
      @Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
      $endgroup$
      – user376343
      Dec 4 '18 at 19:54














    • 1




      $begingroup$
      So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:52










    • $begingroup$
      You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
      $endgroup$
      – Federico
      Dec 4 '18 at 19:53










    • $begingroup$
      @Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
      $endgroup$
      – user376343
      Dec 4 '18 at 19:54








    1




    1




    $begingroup$
    So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:52




    $begingroup$
    So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:52












    $begingroup$
    You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
    $endgroup$
    – Federico
    Dec 4 '18 at 19:53




    $begingroup$
    You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
    $endgroup$
    – Federico
    Dec 4 '18 at 19:53












    $begingroup$
    @Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
    $endgroup$
    – user376343
    Dec 4 '18 at 19:54




    $begingroup$
    @Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
    $endgroup$
    – user376343
    Dec 4 '18 at 19:54











    1












    $begingroup$

    A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :



    $$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$



    If you miss on that, calculate it the explicit way, by :



    $$|lambda I - A| =
    begin{vmatrix}
    lambda - 2 & -1 & 0 \
    0 & lambda - 2 & 0 \
    0 & 0 & lambda - 2 \
    end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$



    $$Rightarrow$$



    $$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$



    The eigenvalues then, would simply be :



    $$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:59










    • $begingroup$
      @EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
      $endgroup$
      – Rebellos
      Dec 4 '18 at 20:03


















    1












    $begingroup$

    A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :



    $$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$



    If you miss on that, calculate it the explicit way, by :



    $$|lambda I - A| =
    begin{vmatrix}
    lambda - 2 & -1 & 0 \
    0 & lambda - 2 & 0 \
    0 & 0 & lambda - 2 \
    end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$



    $$Rightarrow$$



    $$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$



    The eigenvalues then, would simply be :



    $$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:59










    • $begingroup$
      @EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
      $endgroup$
      – Rebellos
      Dec 4 '18 at 20:03
















    1












    1








    1





    $begingroup$

    A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :



    $$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$



    If you miss on that, calculate it the explicit way, by :



    $$|lambda I - A| =
    begin{vmatrix}
    lambda - 2 & -1 & 0 \
    0 & lambda - 2 & 0 \
    0 & 0 & lambda - 2 \
    end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$



    $$Rightarrow$$



    $$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$



    The eigenvalues then, would simply be :



    $$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$






    share|cite|improve this answer











    $endgroup$



    A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :



    $$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$



    If you miss on that, calculate it the explicit way, by :



    $$|lambda I - A| =
    begin{vmatrix}
    lambda - 2 & -1 & 0 \
    0 & lambda - 2 & 0 \
    0 & 0 & lambda - 2 \
    end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$



    $$Rightarrow$$



    $$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$



    The eigenvalues then, would simply be :



    $$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 4 '18 at 19:59

























    answered Dec 4 '18 at 19:44









    RebellosRebellos

    14.5k31246




    14.5k31246












    • $begingroup$
      Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:59










    • $begingroup$
      @EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
      $endgroup$
      – Rebellos
      Dec 4 '18 at 20:03




















    • $begingroup$
      Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
      $endgroup$
      – Evan Kim
      Dec 4 '18 at 19:59










    • $begingroup$
      @EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
      $endgroup$
      – Rebellos
      Dec 4 '18 at 20:03


















    $begingroup$
    Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:59




    $begingroup$
    Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 19:59












    $begingroup$
    @EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
    $endgroup$
    – Rebellos
    Dec 4 '18 at 20:03






    $begingroup$
    @EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
    $endgroup$
    – Rebellos
    Dec 4 '18 at 20:03













    0












    $begingroup$

    In fact, your matrix is special. it makes the computation easier.



    Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
    $$(x,y,z)ne (0,0,0)$$
    and
    $$2x+y=kx$$
    $$2y=ky$$
    $$2z=kz$$
    but
    $$kne 2implies z=y=x=0$$



    which is in contradiction with the assumption.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      In fact, your matrix is special. it makes the computation easier.



      Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
      $$(x,y,z)ne (0,0,0)$$
      and
      $$2x+y=kx$$
      $$2y=ky$$
      $$2z=kz$$
      but
      $$kne 2implies z=y=x=0$$



      which is in contradiction with the assumption.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        In fact, your matrix is special. it makes the computation easier.



        Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
        $$(x,y,z)ne (0,0,0)$$
        and
        $$2x+y=kx$$
        $$2y=ky$$
        $$2z=kz$$
        but
        $$kne 2implies z=y=x=0$$



        which is in contradiction with the assumption.






        share|cite|improve this answer









        $endgroup$



        In fact, your matrix is special. it makes the computation easier.



        Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
        $$(x,y,z)ne (0,0,0)$$
        and
        $$2x+y=kx$$
        $$2y=ky$$
        $$2z=kz$$
        but
        $$kne 2implies z=y=x=0$$



        which is in contradiction with the assumption.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 19:46









        hamam_Abdallahhamam_Abdallah

        38k21634




        38k21634






























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