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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3100338%2feigenvalues-of-a-matrix-whose-square-is-zero%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    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
















    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3100338%2feigenvalues-of-a-matrix-whose-square-is-zero%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            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]
            Read more

            File:DeusFollowingSea.jpg

            Image
            File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
            Read more