Find all of the eigenvalues and eigenvectors of the 2x2 matrix

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$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$



-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next







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    Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
    – cdipaolo
    Aug 9 at 19:30






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    For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
    – JMoravitz
    Aug 9 at 19:30














up vote
1
down vote

favorite












$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$



-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next







share|cite|improve this question


















  • 1




    Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
    – cdipaolo
    Aug 9 at 19:30






  • 1




    For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
    – JMoravitz
    Aug 9 at 19:30












up vote
1
down vote

favorite









up vote
1
down vote

favorite











$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$



-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next







share|cite|improve this question














$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
I got $λ^2 - 3λ - 4 + 6i^2 = 0$



-I isoloated down to $i^2 = -1/6 (λ -4)(λ+1)$ and wrote how λ must be -1 < λ < 4. I don't know what to do next









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 9 at 19:30









mrtaurho

696219




696219










asked Aug 9 at 19:25









Kayy Wang

63




63







  • 1




    Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
    – cdipaolo
    Aug 9 at 19:30






  • 1




    For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
    – JMoravitz
    Aug 9 at 19:30












  • 1




    Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
    – cdipaolo
    Aug 9 at 19:30






  • 1




    For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
    – JMoravitz
    Aug 9 at 19:30







1




1




Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30




Do you know that $i^2=-1$? That’ll give you a polynomial $lambda^2 -3lambda -10=0$ which I assume you can solve.
– cdipaolo
Aug 9 at 19:30




1




1




For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30




For future reference, this page gives information on how to type mathematical equations and such on this site using MathJax and $LaTeX$, making your posts much more readable.
– JMoravitz
Aug 9 at 19:30










2 Answers
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Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?






share|cite|improve this answer




















  • Thank you very much
    – Kayy Wang
    Aug 9 at 19:35










  • You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
    – mrtaurho
    Aug 9 at 19:37

















up vote
1
down vote













$$A = beginpmatrix1&2i\
-3i&-4endpmatrix$$
Now $|A-lambda I|=0$
$$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
$$(1-lambda)(-4-lambda)+6i^2=0$$
$$-4-lambda+4lambda+lambda^2-6=0$$
$$lambda^2+3lambda-10=0$$
$$(lambda+5)(lambda-2)$$
$$lambda=-5,2$$



To find the eigenvector you need to find the null space using the above eigen values






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    2 Answers
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    active

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    2 Answers
    2






    active

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    up vote
    1
    down vote













    Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?






    share|cite|improve this answer




















    • Thank you very much
      – Kayy Wang
      Aug 9 at 19:35










    • You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
      – mrtaurho
      Aug 9 at 19:37














    up vote
    1
    down vote













    Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?






    share|cite|improve this answer




















    • Thank you very much
      – Kayy Wang
      Aug 9 at 19:35










    • You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
      – mrtaurho
      Aug 9 at 19:37












    up vote
    1
    down vote










    up vote
    1
    down vote









    Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?






    share|cite|improve this answer












    Just use the fact that $i^2=-1$ within your determinate of the matrix. Furthermore it should be $+3lambda$ instead of $-3lambda$. Therefore you got $lambda^2+3lambda-4+6(-1)=0$. Can you proceed from here?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 9 at 19:29









    mrtaurho

    696219




    696219











    • Thank you very much
      – Kayy Wang
      Aug 9 at 19:35










    • You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
      – mrtaurho
      Aug 9 at 19:37
















    • Thank you very much
      – Kayy Wang
      Aug 9 at 19:35










    • You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
      – mrtaurho
      Aug 9 at 19:37















    Thank you very much
    – Kayy Wang
    Aug 9 at 19:35




    Thank you very much
    – Kayy Wang
    Aug 9 at 19:35












    You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
    – mrtaurho
    Aug 9 at 19:37




    You are welcome. Just accept the answer to show that you are satisfied with the answers to your post :)
    – mrtaurho
    Aug 9 at 19:37










    up vote
    1
    down vote













    $$A = beginpmatrix1&2i\
    -3i&-4endpmatrix$$
    Now $|A-lambda I|=0$
    $$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
    $$(1-lambda)(-4-lambda)+6i^2=0$$
    $$-4-lambda+4lambda+lambda^2-6=0$$
    $$lambda^2+3lambda-10=0$$
    $$(lambda+5)(lambda-2)$$
    $$lambda=-5,2$$



    To find the eigenvector you need to find the null space using the above eigen values






    share|cite|improve this answer
























      up vote
      1
      down vote













      $$A = beginpmatrix1&2i\
      -3i&-4endpmatrix$$
      Now $|A-lambda I|=0$
      $$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
      $$(1-lambda)(-4-lambda)+6i^2=0$$
      $$-4-lambda+4lambda+lambda^2-6=0$$
      $$lambda^2+3lambda-10=0$$
      $$(lambda+5)(lambda-2)$$
      $$lambda=-5,2$$



      To find the eigenvector you need to find the null space using the above eigen values






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        $$A = beginpmatrix1&2i\
        -3i&-4endpmatrix$$
        Now $|A-lambda I|=0$
        $$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
        $$(1-lambda)(-4-lambda)+6i^2=0$$
        $$-4-lambda+4lambda+lambda^2-6=0$$
        $$lambda^2+3lambda-10=0$$
        $$(lambda+5)(lambda-2)$$
        $$lambda=-5,2$$



        To find the eigenvector you need to find the null space using the above eigen values






        share|cite|improve this answer












        $$A = beginpmatrix1&2i\
        -3i&-4endpmatrix$$
        Now $|A-lambda I|=0$
        $$beginvmatrix 1-lambda & 2i \ -3i & -4-lambda endvmatrix=0$$
        $$(1-lambda)(-4-lambda)+6i^2=0$$
        $$-4-lambda+4lambda+lambda^2-6=0$$
        $$lambda^2+3lambda-10=0$$
        $$(lambda+5)(lambda-2)$$
        $$lambda=-5,2$$



        To find the eigenvector you need to find the null space using the above eigen values







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 9 at 19:34









        Key Flex

        4,863628




        4,863628






















             

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