Find $2times 2$ symmetric matrix $A$ given two eigenvalues and one eigenvector












1












$begingroup$


I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04


















1












$begingroup$


I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04
















1












1








1





$begingroup$


I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 22:06









user376343

3,4033826




3,4033826










asked Dec 7 '18 at 21:52









a hea he

61




61








  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04
















  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04










3




3




$begingroup$
Different eigenspaces of a symmetric matrix are orthogonal to each other.
$endgroup$
– Berci
Dec 7 '18 at 21:54




$begingroup$
Different eigenspaces of a symmetric matrix are orthogonal to each other.
$endgroup$
– Berci
Dec 7 '18 at 21:54




1




1




$begingroup$
...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
$endgroup$
– Jean Marie
Dec 7 '18 at 22:04






$begingroup$
...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
$endgroup$
– Jean Marie
Dec 7 '18 at 22:04












2 Answers
2






active

oldest

votes


















2












$begingroup$

To offer a slightly different perspective, due to the Spectral Theorem you have
$$
A=P_1+4P_2,$$
where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
$$
P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
$$
And then
$$
A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let :



    $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



    Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



    $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



    such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



    $$(a-lambda)(d-lambda)-bc=0$$



    Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



    $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



    $$Leftrightarrow$$



    $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



    Can you now find $a,b,c$ and $d$ ?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
      $endgroup$
      – M. A. SARKAR
      Dec 7 '18 at 22:10













    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%2f3030404%2ffind-2-times-2-symmetric-matrix-a-given-two-eigenvalues-and-one-eigenvector%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    To offer a slightly different perspective, due to the Spectral Theorem you have
    $$
    A=P_1+4P_2,$$
    where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
    $$
    P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
    $$
    And then
    $$
    A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
    $$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      To offer a slightly different perspective, due to the Spectral Theorem you have
      $$
      A=P_1+4P_2,$$
      where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
      $$
      P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
      $$
      And then
      $$
      A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
      $$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        To offer a slightly different perspective, due to the Spectral Theorem you have
        $$
        A=P_1+4P_2,$$
        where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
        $$
        P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
        $$
        And then
        $$
        A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
        $$






        share|cite|improve this answer









        $endgroup$



        To offer a slightly different perspective, due to the Spectral Theorem you have
        $$
        A=P_1+4P_2,$$
        where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
        $$
        P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
        $$
        And then
        $$
        A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 22:12









        Martin ArgeramiMartin Argerami

        126k1182180




        126k1182180























            1












            $begingroup$

            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10


















            1












            $begingroup$

            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10
















            1












            1








            1





            $begingroup$

            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?






            share|cite|improve this answer









            $endgroup$



            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 7 '18 at 22:07









            RebellosRebellos

            14.6k31247




            14.6k31247












            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10




















            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10


















            $begingroup$
            Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
            $endgroup$
            – M. A. SARKAR
            Dec 7 '18 at 22:10






            $begingroup$
            Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
            $endgroup$
            – M. A. SARKAR
            Dec 7 '18 at 22:10




















            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%2f3030404%2ffind-2-times-2-symmetric-matrix-a-given-two-eigenvalues-and-one-eigenvector%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

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more