Eigenvalues of a matrix whose square is zero












3












$begingroup$


Let $A$ be a nonzero 3×3 matrix such that $A^2=0$. Then what is the number of non-zero eigenvalues of the matrix?
I am unable to figure out the eigenvalues of the above matrix.
P.S.: how would the answer change if it were given that $A^3=0$?






linear-algebra eigenvalues-eigenvectors







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  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    5 hours ago
















3












$begingroup$


Let $A$ be a nonzero 3×3 matrix such that $A^2=0$. Then what is the number of non-zero eigenvalues of the matrix?
I am unable to figure out the eigenvalues of the above matrix.
P.S.: how would the answer change if it were given that $A^3=0$?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    5 hours ago














3












3








3





$begingroup$


Let $A$ be a nonzero 3×3 matrix such that $A^2=0$. Then what is the number of non-zero eigenvalues of the matrix?
I am unable to figure out the eigenvalues of the above matrix.
P.S.: how would the answer change if it were given that $A^3=0$?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




Let $A$ be a nonzero 3×3 matrix such that $A^2=0$. Then what is the number of non-zero eigenvalues of the matrix?
I am unable to figure out the eigenvalues of the above matrix.
P.S.: how would the answer change if it were given that $A^3=0$?






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited 3 hours ago









J. W. Tanner

1,306114




1,306114










asked 5 hours ago









Jor_ElJor_El

356




356












  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    5 hours ago


















  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    5 hours ago
















$begingroup$
Hint: Eigenvalues are roots of the characteristic polynomial
$endgroup$
– ab123
5 hours ago




$begingroup$
Hint: Eigenvalues are roots of the characteristic polynomial
$endgroup$
– ab123
5 hours ago










3 Answers
3






active

oldest

votes


















9












$begingroup$

A square matrix $A$ is called nilpotent if there is a $p in mathbb{N}$ such that $A^p=0$. So let $A$ be a nilpotent matrix. Then we have by definition of an eigenvalue



$$Av=lambda v,$$



where $lambda$ is an eigenvalue of $A$ and $vneq 0$ is an eigenvector of $A$ to the corresponding eigenvalue. Because $A$ is nilpotent we also have



$$0=A^p v=lambda^p v$$



and because $v neq 0$ it follows $lambda^p=0$, i. e. $lambda=0$. So to your question: The number of non zero eigenvalues is in this case $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So does it mean that a nilpotent matrix has all eigen values equal to 0?
    $endgroup$
    – Jor_El
    5 hours ago










  • $begingroup$
    @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
    $endgroup$
    – Jan
    5 hours ago



















2












$begingroup$

Another approach is this one:



Since $A^2 = 0$, the polynomial $g(x) = x^2$ annihilates A (and this means that the Linear operator defined by $g(T)$ is the null operator). However, the minimal polynomial of A must divide every polynomial that annihilates $A$, so if $m(x)$ is such polynomial, $m$ must divide $g$.



Hence, $m(x) = x^2$, because $A ≠ 0$.



Thus, the characteristical polynomial of $A$ is $p(x) = x^2$, because $p(x)$ has the same roots of $m(x)$ (why?), and $p(x)$ annihilates $A$ (by Cayley-Hamilton theorem).



In conclusion, the characteristical polynomial of $A$ has only a single root, $0$, and since every eigenvalue of $A$ is a root of it's characteristical polynomial, we have that 0 is the only eigenvalue of $A$.






share|cite|improve this answer








New contributor




Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension.
    $endgroup$
    – Mike Earnest
    1 hour ago










  • $begingroup$
    More generally, if some linear operator $B$ has $Bv = lambda v$ with $vne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $lambda$ of $B$ must be a root of $P$.
    $endgroup$
    – Paul Sinclair
    26 mins ago





















1












$begingroup$

Clearly, as the characteristic polynomial of the matrix $A$ is $x^n = 0$, and the eigenvalues are roots of the characteristic polynomial, there can not be any non-zero eigenvalue.






share|cite|improve this answer









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    3 Answers
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    active

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    3 Answers
    3






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    active

    oldest

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    active

    oldest

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    9












    $begingroup$

    A square matrix $A$ is called nilpotent if there is a $p in mathbb{N}$ such that $A^p=0$. So let $A$ be a nilpotent matrix. Then we have by definition of an eigenvalue



    $$Av=lambda v,$$



    where $lambda$ is an eigenvalue of $A$ and $vneq 0$ is an eigenvector of $A$ to the corresponding eigenvalue. Because $A$ is nilpotent we also have



    $$0=A^p v=lambda^p v$$



    and because $v neq 0$ it follows $lambda^p=0$, i. e. $lambda=0$. So to your question: The number of non zero eigenvalues is in this case $0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So does it mean that a nilpotent matrix has all eigen values equal to 0?
      $endgroup$
      – Jor_El
      5 hours ago










    • $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      5 hours ago
















    9












    $begingroup$

    A square matrix $A$ is called nilpotent if there is a $p in mathbb{N}$ such that $A^p=0$. So let $A$ be a nilpotent matrix. Then we have by definition of an eigenvalue



    $$Av=lambda v,$$



    where $lambda$ is an eigenvalue of $A$ and $vneq 0$ is an eigenvector of $A$ to the corresponding eigenvalue. Because $A$ is nilpotent we also have



    $$0=A^p v=lambda^p v$$



    and because $v neq 0$ it follows $lambda^p=0$, i. e. $lambda=0$. So to your question: The number of non zero eigenvalues is in this case $0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So does it mean that a nilpotent matrix has all eigen values equal to 0?
      $endgroup$
      – Jor_El
      5 hours ago










    • $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      5 hours ago














    9












    9








    9





    $begingroup$

    A square matrix $A$ is called nilpotent if there is a $p in mathbb{N}$ such that $A^p=0$. So let $A$ be a nilpotent matrix. Then we have by definition of an eigenvalue



    $$Av=lambda v,$$



    where $lambda$ is an eigenvalue of $A$ and $vneq 0$ is an eigenvector of $A$ to the corresponding eigenvalue. Because $A$ is nilpotent we also have



    $$0=A^p v=lambda^p v$$



    and because $v neq 0$ it follows $lambda^p=0$, i. e. $lambda=0$. So to your question: The number of non zero eigenvalues is in this case $0$.






    share|cite|improve this answer









    $endgroup$



    A square matrix $A$ is called nilpotent if there is a $p in mathbb{N}$ such that $A^p=0$. So let $A$ be a nilpotent matrix. Then we have by definition of an eigenvalue



    $$Av=lambda v,$$



    where $lambda$ is an eigenvalue of $A$ and $vneq 0$ is an eigenvector of $A$ to the corresponding eigenvalue. Because $A$ is nilpotent we also have



    $$0=A^p v=lambda^p v$$



    and because $v neq 0$ it follows $lambda^p=0$, i. e. $lambda=0$. So to your question: The number of non zero eigenvalues is in this case $0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 5 hours ago









    JanJan

    331114




    331114












    • $begingroup$
      So does it mean that a nilpotent matrix has all eigen values equal to 0?
      $endgroup$
      – Jor_El
      5 hours ago










    • $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      5 hours ago


















    • $begingroup$
      So does it mean that a nilpotent matrix has all eigen values equal to 0?
      $endgroup$
      – Jor_El
      5 hours ago










    • $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      5 hours ago
















    $begingroup$
    So does it mean that a nilpotent matrix has all eigen values equal to 0?
    $endgroup$
    – Jor_El
    5 hours ago




    $begingroup$
    So does it mean that a nilpotent matrix has all eigen values equal to 0?
    $endgroup$
    – Jor_El
    5 hours ago












    $begingroup$
    @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
    $endgroup$
    – Jan
    5 hours ago




    $begingroup$
    @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
    $endgroup$
    – Jan
    5 hours ago











    2












    $begingroup$

    Another approach is this one:



    Since $A^2 = 0$, the polynomial $g(x) = x^2$ annihilates A (and this means that the Linear operator defined by $g(T)$ is the null operator). However, the minimal polynomial of A must divide every polynomial that annihilates $A$, so if $m(x)$ is such polynomial, $m$ must divide $g$.



    Hence, $m(x) = x^2$, because $A ≠ 0$.



    Thus, the characteristical polynomial of $A$ is $p(x) = x^2$, because $p(x)$ has the same roots of $m(x)$ (why?), and $p(x)$ annihilates $A$ (by Cayley-Hamilton theorem).



    In conclusion, the characteristical polynomial of $A$ has only a single root, $0$, and since every eigenvalue of $A$ is a root of it's characteristical polynomial, we have that 0 is the only eigenvalue of $A$.






    share|cite|improve this answer








    New contributor




    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension.
      $endgroup$
      – Mike Earnest
      1 hour ago










    • $begingroup$
      More generally, if some linear operator $B$ has $Bv = lambda v$ with $vne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $lambda$ of $B$ must be a root of $P$.
      $endgroup$
      – Paul Sinclair
      26 mins ago


















    2












    $begingroup$

    Another approach is this one:



    Since $A^2 = 0$, the polynomial $g(x) = x^2$ annihilates A (and this means that the Linear operator defined by $g(T)$ is the null operator). However, the minimal polynomial of A must divide every polynomial that annihilates $A$, so if $m(x)$ is such polynomial, $m$ must divide $g$.



    Hence, $m(x) = x^2$, because $A ≠ 0$.



    Thus, the characteristical polynomial of $A$ is $p(x) = x^2$, because $p(x)$ has the same roots of $m(x)$ (why?), and $p(x)$ annihilates $A$ (by Cayley-Hamilton theorem).



    In conclusion, the characteristical polynomial of $A$ has only a single root, $0$, and since every eigenvalue of $A$ is a root of it's characteristical polynomial, we have that 0 is the only eigenvalue of $A$.






    share|cite|improve this answer








    New contributor




    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension.
      $endgroup$
      – Mike Earnest
      1 hour ago










    • $begingroup$
      More generally, if some linear operator $B$ has $Bv = lambda v$ with $vne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $lambda$ of $B$ must be a root of $P$.
      $endgroup$
      – Paul Sinclair
      26 mins ago
















    2












    2








    2





    $begingroup$

    Another approach is this one:



    Since $A^2 = 0$, the polynomial $g(x) = x^2$ annihilates A (and this means that the Linear operator defined by $g(T)$ is the null operator). However, the minimal polynomial of A must divide every polynomial that annihilates $A$, so if $m(x)$ is such polynomial, $m$ must divide $g$.



    Hence, $m(x) = x^2$, because $A ≠ 0$.



    Thus, the characteristical polynomial of $A$ is $p(x) = x^2$, because $p(x)$ has the same roots of $m(x)$ (why?), and $p(x)$ annihilates $A$ (by Cayley-Hamilton theorem).



    In conclusion, the characteristical polynomial of $A$ has only a single root, $0$, and since every eigenvalue of $A$ is a root of it's characteristical polynomial, we have that 0 is the only eigenvalue of $A$.






    share|cite|improve this answer








    New contributor




    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$



    Another approach is this one:



    Since $A^2 = 0$, the polynomial $g(x) = x^2$ annihilates A (and this means that the Linear operator defined by $g(T)$ is the null operator). However, the minimal polynomial of A must divide every polynomial that annihilates $A$, so if $m(x)$ is such polynomial, $m$ must divide $g$.



    Hence, $m(x) = x^2$, because $A ≠ 0$.



    Thus, the characteristical polynomial of $A$ is $p(x) = x^2$, because $p(x)$ has the same roots of $m(x)$ (why?), and $p(x)$ annihilates $A$ (by Cayley-Hamilton theorem).



    In conclusion, the characteristical polynomial of $A$ has only a single root, $0$, and since every eigenvalue of $A$ is a root of it's characteristical polynomial, we have that 0 is the only eigenvalue of $A$.







    share|cite|improve this answer








    New contributor




    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer






    New contributor




    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered 3 hours ago









    Bruno TassoneBruno Tassone

    213




    213




    New contributor




    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





    New contributor





    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    Bruno Tassone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.












    • $begingroup$
      One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension.
      $endgroup$
      – Mike Earnest
      1 hour ago










    • $begingroup$
      More generally, if some linear operator $B$ has $Bv = lambda v$ with $vne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $lambda$ of $B$ must be a root of $P$.
      $endgroup$
      – Paul Sinclair
      26 mins ago




















    • $begingroup$
      One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension.
      $endgroup$
      – Mike Earnest
      1 hour ago










    • $begingroup$
      More generally, if some linear operator $B$ has $Bv = lambda v$ with $vne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $lambda$ of $B$ must be a root of $P$.
      $endgroup$
      – Paul Sinclair
      26 mins ago


















    $begingroup$
    One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension.
    $endgroup$
    – Mike Earnest
    1 hour ago




    $begingroup$
    One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension.
    $endgroup$
    – Mike Earnest
    1 hour ago












    $begingroup$
    More generally, if some linear operator $B$ has $Bv = lambda v$ with $vne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $lambda$ of $B$ must be a root of $P$.
    $endgroup$
    – Paul Sinclair
    26 mins ago






    $begingroup$
    More generally, if some linear operator $B$ has $Bv = lambda v$ with $vne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $lambda$ of $B$ must be a root of $P$.
    $endgroup$
    – Paul Sinclair
    26 mins ago













    1












    $begingroup$

    Clearly, as the characteristic polynomial of the matrix $A$ is $x^n = 0$, and the eigenvalues are roots of the characteristic polynomial, there can not be any non-zero eigenvalue.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Clearly, as the characteristic polynomial of the matrix $A$ is $x^n = 0$, and the eigenvalues are roots of the characteristic polynomial, there can not be any non-zero eigenvalue.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Clearly, as the characteristic polynomial of the matrix $A$ is $x^n = 0$, and the eigenvalues are roots of the characteristic polynomial, there can not be any non-zero eigenvalue.






        share|cite|improve this answer









        $endgroup$



        Clearly, as the characteristic polynomial of the matrix $A$ is $x^n = 0$, and the eigenvalues are roots of the characteristic polynomial, there can not be any non-zero eigenvalue.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        ab123ab123

        1,781423




        1,781423






























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