Eigenvalues of a matrix whose square is zero












6












$begingroup$


Let $A$ be a nonzero $3 times 3$ 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 matrices eigenvalues-eigenvectors matrix-equations nilpotence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    Feb 4 at 20:05
















6












$begingroup$


Let $A$ be a nonzero $3 times 3$ 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 matrices eigenvalues-eigenvectors matrix-equations nilpotence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    Feb 4 at 20:05














6












6








6





$begingroup$


Let $A$ be a nonzero $3 times 3$ 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 matrices eigenvalues-eigenvectors matrix-equations nilpotence







share|cite|improve this question











$endgroup$




Let $A$ be a nonzero $3 times 3$ 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 matrices eigenvalues-eigenvectors matrix-equations nilpotence




linear-algebra matrices eigenvalues-eigenvectors matrix-equations nilpotence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 5 at 8:37









Rodrigo de Azevedo

13.2k41960




13.2k41960










asked Feb 4 at 20:00









Jor_ElJor_El

696




696












  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    Feb 4 at 20:05


















  • $begingroup$
    Hint: Eigenvalues are roots of the characteristic polynomial
    $endgroup$
    – ab123
    Feb 4 at 20:05
















$begingroup$
Hint: Eigenvalues are roots of the characteristic polynomial
$endgroup$
– ab123
Feb 4 at 20:05




$begingroup$
Hint: Eigenvalues are roots of the characteristic polynomial
$endgroup$
– ab123
Feb 4 at 20:05










3 Answers
3






active

oldest

votes


















12












$begingroup$

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
    Feb 4 at 20:11






  • 1




    $begingroup$
    @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
    $endgroup$
    – Jan
    Feb 4 at 20:13










  • $begingroup$
    As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity.
    $endgroup$
    – Michael Anderson
    Feb 5 at 1:53



















2












$begingroup$

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









$endgroup$





















    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 $p(x) = x^3$, 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











    $endgroup$









    • 2




      $begingroup$
      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
      Feb 4 at 23:39










    • $begingroup$
      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
      Feb 5 at 0:57












    • $begingroup$
      You're right. I'll fix it right away.
      $endgroup$
      – Bruno Tassone
      Feb 5 at 8:56











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    3 Answers
    3






    active

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    3 Answers
    3






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    active

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    active

    oldest

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    12












    $begingroup$

    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
      Feb 4 at 20:11






    • 1




      $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      Feb 4 at 20:13










    • $begingroup$
      As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity.
      $endgroup$
      – Michael Anderson
      Feb 5 at 1:53
















    12












    $begingroup$

    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
      Feb 4 at 20:11






    • 1




      $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      Feb 4 at 20:13










    • $begingroup$
      As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity.
      $endgroup$
      – Michael Anderson
      Feb 5 at 1:53














    12












    12








    12





    $begingroup$

    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 Feb 4 at 20:06









    JanJan

    605315




    605315












    • $begingroup$
      So does it mean that a nilpotent matrix has all eigen values equal to 0?
      $endgroup$
      – Jor_El
      Feb 4 at 20:11






    • 1




      $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      Feb 4 at 20:13










    • $begingroup$
      As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity.
      $endgroup$
      – Michael Anderson
      Feb 5 at 1:53


















    • $begingroup$
      So does it mean that a nilpotent matrix has all eigen values equal to 0?
      $endgroup$
      – Jor_El
      Feb 4 at 20:11






    • 1




      $begingroup$
      @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
      $endgroup$
      – Jan
      Feb 4 at 20:13










    • $begingroup$
      As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity.
      $endgroup$
      – Michael Anderson
      Feb 5 at 1:53
















    $begingroup$
    So does it mean that a nilpotent matrix has all eigen values equal to 0?
    $endgroup$
    – Jor_El
    Feb 4 at 20:11




    $begingroup$
    So does it mean that a nilpotent matrix has all eigen values equal to 0?
    $endgroup$
    – Jor_El
    Feb 4 at 20:11




    1




    1




    $begingroup$
    @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
    $endgroup$
    – Jan
    Feb 4 at 20:13




    $begingroup$
    @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero.
    $endgroup$
    – Jan
    Feb 4 at 20:13












    $begingroup$
    As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity.
    $endgroup$
    – Michael Anderson
    Feb 5 at 1:53




    $begingroup$
    As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity.
    $endgroup$
    – Michael Anderson
    Feb 5 at 1:53











    2












    $begingroup$

    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









    $endgroup$


















      2












      $begingroup$

      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









      $endgroup$
















        2












        2








        2





        $begingroup$

        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 Feb 4 at 20:09









        ab123ab123

        1,799423




        1,799423























            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 $p(x) = x^3$, 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











            $endgroup$









            • 2




              $begingroup$
              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
              Feb 4 at 23:39










            • $begingroup$
              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
              Feb 5 at 0:57












            • $begingroup$
              You're right. I'll fix it right away.
              $endgroup$
              – Bruno Tassone
              Feb 5 at 8:56
















            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 $p(x) = x^3$, 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











            $endgroup$









            • 2




              $begingroup$
              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
              Feb 4 at 23:39










            • $begingroup$
              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
              Feb 5 at 0:57












            • $begingroup$
              You're right. I'll fix it right away.
              $endgroup$
              – Bruno Tassone
              Feb 5 at 8:56














            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 $p(x) = x^3$, 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











            $endgroup$



            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 $p(x) = x^3$, 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














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 5 at 8:57

























            answered Feb 4 at 21:28









            Bruno TassoneBruno Tassone

            687




            687








            • 2




              $begingroup$
              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
              Feb 4 at 23:39










            • $begingroup$
              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
              Feb 5 at 0:57












            • $begingroup$
              You're right. I'll fix it right away.
              $endgroup$
              – Bruno Tassone
              Feb 5 at 8:56














            • 2




              $begingroup$
              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
              Feb 4 at 23:39










            • $begingroup$
              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
              Feb 5 at 0:57












            • $begingroup$
              You're right. I'll fix it right away.
              $endgroup$
              – Bruno Tassone
              Feb 5 at 8:56








            2




            2




            $begingroup$
            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
            Feb 4 at 23:39




            $begingroup$
            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
            Feb 4 at 23:39












            $begingroup$
            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
            Feb 5 at 0:57






            $begingroup$
            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
            Feb 5 at 0:57














            $begingroup$
            You're right. I'll fix it right away.
            $endgroup$
            – Bruno Tassone
            Feb 5 at 8:56




            $begingroup$
            You're right. I'll fix it right away.
            $endgroup$
            – Bruno Tassone
            Feb 5 at 8:56


















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            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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