How can i solve the eigenvector equation for a $2 times 2$ matrix with eigenvalues $lambda = -1$ and...












0












$begingroup$


where



$$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



(For clarification, the materials i am using are telling me to use the following formula and to input the appropriate values for a, b, c and d but i am getting terribly confused.)



(a-λ)x+by=0



cx+(d-λ)y=0






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    where



    $$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



    (For clarification, the materials i am using are telling me to use the following formula and to input the appropriate values for a, b, c and d but i am getting terribly confused.)



    (a-λ)x+by=0



    cx+(d-λ)y=0






    linear-algebra eigenvalues-eigenvectors







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      where



      $$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



      (For clarification, the materials i am using are telling me to use the following formula and to input the appropriate values for a, b, c and d but i am getting terribly confused.)



      (a-λ)x+by=0



      cx+(d-λ)y=0






      linear-algebra eigenvalues-eigenvectors







      share|cite|improve this question











      $endgroup$




      where



      $$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



      (For clarification, the materials i am using are telling me to use the following formula and to input the appropriate values for a, b, c and d but i am getting terribly confused.)



      (a-λ)x+by=0



      cx+(d-λ)y=0






      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 Jan 2 at 20:51







      RocketKangaroo

















      asked Jan 2 at 19:19









      RocketKangarooRocketKangaroo

      334




      334






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          It's pretty straightforward; I'll work out the case $lambda = 3$; suppose



          $vec v = begin{pmatrix} a \ b end{pmatrix} tag 1$



          is the eigenvector corresponding to $lambda = 3$; then



          $A vec v = 3vec v; tag 2$



          we write this equation out using the given entries of $A$, $vec v$:



          $begin{bmatrix} 1 & 1 \ 4 & 1 end{bmatrix} begin{pmatrix} a \ b end{pmatrix} = 3 begin{pmatrix} a \ b end{pmatrix} = begin{pmatrix} 3a \ 3b end{pmatrix}, tag 3$



          or



          $a + b = 3a, tag 4$



          $4a + b = 3b; tag 5$



          from (4),



          $b = 2a; tag 6$



          it is expeditious at this point to observe we cannot have an eigenvector with $a = 0$, lest by (6) $b = 0$ as well, and thus



          $vec v = 0; tag 7$



          but eigenvectors are non-vanishing by definition. Thus



          $a ne 0, tag 8$



          and since eigenvectors are scalable, that is, the quality of "eigenvector-ness" is invariant under scalar multiplication, we may assume



          $a = 1; tag 9$



          then



          $b = 2a = 2, tag{10}$



          $vec v = begin{pmatrix} 1 \ 2 end{pmatrix}; tag{11}$



          it is easy to see that



          $4a + b = 4(1) + 2 = 6 = 3(2), tag{12}$



          so $vec v$ also satisfies (5).



          That's about it for the case $lambda = 3$; I leave the case $lambda = -1$ to the reader.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Hi, could you expand on how you went from (3) to (6) please? I'm getting confused because the materials i am working from are telling me to use the following formula for a 2x2 matrix; (a-λ)x+by=0 and cx+(d-λ)y=0.
            $endgroup$
            – RocketKangaroo
            Jan 2 at 20:49












          • $begingroup$
            @RocketKangaroo: look at (4), which is just the first row of (3) written out. (4) says $a + b = 3a$; now subtract $a$ from both sides; you get (6), OK? The formula $(a - lambda)x + by = 0$ is also the first row of $Avec w = lambda vec w$ written out, with $A = begin{bmatrix} a & b \ c & d end{bmatrix}$ and $vec w = (x, y)^T$; from $A vec w = lambda vec w$ we get $ax + by = lambda a$ or $(a - lambda)x + by = 0$. It's just the general case of what I did. Does this help?
            $endgroup$
            – Robert Lewis
            Jan 2 at 20:57












          • $begingroup$
            Yes indeed, i'm still a little confused on how you figured out a = 1 however, is this because 0 cannot exist as an eigenvector?
            $endgroup$
            – RocketKangaroo
            Jan 2 at 21:25










          • $begingroup$
            @RocketKangaroo: see (6)-(7) and the words between; they explain why $a ne 0$; then we just multiply $vec v$ by $a^{-1}$, whatever $a ne 0$ may be.
            $endgroup$
            – Robert Lewis
            Jan 2 at 21:31










          • $begingroup$
            @RocketKangaroo: yes, because $0$ cannot be an eigenvector.
            $endgroup$
            – Robert Lewis
            Jan 2 at 21:32



















          2












          $begingroup$

          To find an eigenvector of $lambda=3$ you need to solve :
          $$AX = lambda X $$
          $$begin{bmatrix}1&1\4&1end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}=3begin{bmatrix}x_1\x_2end{bmatrix}$$
          $$x_1+x_2 = 3x_1 text{ and }4x_1+x_2=3x_2 $$
          Thus $X=begin{bmatrix}1\2end{bmatrix}$ is an eigenvector for $lambda=3$.



          You can do the same reasonning for $lambda=-1$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Gee, your answer is almost the same as mine! How could that possible be? ;) +1
            $endgroup$
            – Robert Lewis
            Jan 2 at 19:51






          • 1




            $begingroup$
            Yours is way more detailed :)
            $endgroup$
            – Jennifer
            Jan 2 at 19:53










          • $begingroup$
            True enough. Maybe I talk too much! ;)
            $endgroup$
            – Robert Lewis
            Jan 2 at 19:55



















          0












          $begingroup$

          The characteristic polynomial of $A$ is
          $$
          (lambda-1)^2-4=(lambda-1-2)(lambda-1+2)=(lambda-3)(lambda+1)
          $$
          Therefore $(A-3I)(A+I)=(A+I)(A-3I)=0$.
          So the columns of $(A-3I)$ are solutions of $(A+I)X=0$. And the columns of $(A+I)$ are solutions of $(A-3I)X=0$.



          Concretely,
          $$
          A-3I = left[begin{array}{cc}-2 & 1 \ 4 & -2end{array}right]
          $$
          So
          $$
          (A+I)left[begin{array}{c} 1 \ -2 end{array}right]=0
          $$
          And
          $$
          A+I = left[begin{array}{cc} 2 & 1 \ 4 & 2end{array}right]
          $$
          So
          $$
          (A-3I)left[begin{array}{c}1 \ 2end{array}right] = 0.
          $$






          share|cite|improve this answer









          $endgroup$













            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%2f3059856%2fhow-can-i-solve-the-eigenvector-equation-for-a-2-times-2-matrix-with-eigenval%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









            3












            $begingroup$

            It's pretty straightforward; I'll work out the case $lambda = 3$; suppose



            $vec v = begin{pmatrix} a \ b end{pmatrix} tag 1$



            is the eigenvector corresponding to $lambda = 3$; then



            $A vec v = 3vec v; tag 2$



            we write this equation out using the given entries of $A$, $vec v$:



            $begin{bmatrix} 1 & 1 \ 4 & 1 end{bmatrix} begin{pmatrix} a \ b end{pmatrix} = 3 begin{pmatrix} a \ b end{pmatrix} = begin{pmatrix} 3a \ 3b end{pmatrix}, tag 3$



            or



            $a + b = 3a, tag 4$



            $4a + b = 3b; tag 5$



            from (4),



            $b = 2a; tag 6$



            it is expeditious at this point to observe we cannot have an eigenvector with $a = 0$, lest by (6) $b = 0$ as well, and thus



            $vec v = 0; tag 7$



            but eigenvectors are non-vanishing by definition. Thus



            $a ne 0, tag 8$



            and since eigenvectors are scalable, that is, the quality of "eigenvector-ness" is invariant under scalar multiplication, we may assume



            $a = 1; tag 9$



            then



            $b = 2a = 2, tag{10}$



            $vec v = begin{pmatrix} 1 \ 2 end{pmatrix}; tag{11}$



            it is easy to see that



            $4a + b = 4(1) + 2 = 6 = 3(2), tag{12}$



            so $vec v$ also satisfies (5).



            That's about it for the case $lambda = 3$; I leave the case $lambda = -1$ to the reader.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Hi, could you expand on how you went from (3) to (6) please? I'm getting confused because the materials i am working from are telling me to use the following formula for a 2x2 matrix; (a-λ)x+by=0 and cx+(d-λ)y=0.
              $endgroup$
              – RocketKangaroo
              Jan 2 at 20:49












            • $begingroup$
              @RocketKangaroo: look at (4), which is just the first row of (3) written out. (4) says $a + b = 3a$; now subtract $a$ from both sides; you get (6), OK? The formula $(a - lambda)x + by = 0$ is also the first row of $Avec w = lambda vec w$ written out, with $A = begin{bmatrix} a & b \ c & d end{bmatrix}$ and $vec w = (x, y)^T$; from $A vec w = lambda vec w$ we get $ax + by = lambda a$ or $(a - lambda)x + by = 0$. It's just the general case of what I did. Does this help?
              $endgroup$
              – Robert Lewis
              Jan 2 at 20:57












            • $begingroup$
              Yes indeed, i'm still a little confused on how you figured out a = 1 however, is this because 0 cannot exist as an eigenvector?
              $endgroup$
              – RocketKangaroo
              Jan 2 at 21:25










            • $begingroup$
              @RocketKangaroo: see (6)-(7) and the words between; they explain why $a ne 0$; then we just multiply $vec v$ by $a^{-1}$, whatever $a ne 0$ may be.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:31










            • $begingroup$
              @RocketKangaroo: yes, because $0$ cannot be an eigenvector.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:32
















            3












            $begingroup$

            It's pretty straightforward; I'll work out the case $lambda = 3$; suppose



            $vec v = begin{pmatrix} a \ b end{pmatrix} tag 1$



            is the eigenvector corresponding to $lambda = 3$; then



            $A vec v = 3vec v; tag 2$



            we write this equation out using the given entries of $A$, $vec v$:



            $begin{bmatrix} 1 & 1 \ 4 & 1 end{bmatrix} begin{pmatrix} a \ b end{pmatrix} = 3 begin{pmatrix} a \ b end{pmatrix} = begin{pmatrix} 3a \ 3b end{pmatrix}, tag 3$



            or



            $a + b = 3a, tag 4$



            $4a + b = 3b; tag 5$



            from (4),



            $b = 2a; tag 6$



            it is expeditious at this point to observe we cannot have an eigenvector with $a = 0$, lest by (6) $b = 0$ as well, and thus



            $vec v = 0; tag 7$



            but eigenvectors are non-vanishing by definition. Thus



            $a ne 0, tag 8$



            and since eigenvectors are scalable, that is, the quality of "eigenvector-ness" is invariant under scalar multiplication, we may assume



            $a = 1; tag 9$



            then



            $b = 2a = 2, tag{10}$



            $vec v = begin{pmatrix} 1 \ 2 end{pmatrix}; tag{11}$



            it is easy to see that



            $4a + b = 4(1) + 2 = 6 = 3(2), tag{12}$



            so $vec v$ also satisfies (5).



            That's about it for the case $lambda = 3$; I leave the case $lambda = -1$ to the reader.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Hi, could you expand on how you went from (3) to (6) please? I'm getting confused because the materials i am working from are telling me to use the following formula for a 2x2 matrix; (a-λ)x+by=0 and cx+(d-λ)y=0.
              $endgroup$
              – RocketKangaroo
              Jan 2 at 20:49












            • $begingroup$
              @RocketKangaroo: look at (4), which is just the first row of (3) written out. (4) says $a + b = 3a$; now subtract $a$ from both sides; you get (6), OK? The formula $(a - lambda)x + by = 0$ is also the first row of $Avec w = lambda vec w$ written out, with $A = begin{bmatrix} a & b \ c & d end{bmatrix}$ and $vec w = (x, y)^T$; from $A vec w = lambda vec w$ we get $ax + by = lambda a$ or $(a - lambda)x + by = 0$. It's just the general case of what I did. Does this help?
              $endgroup$
              – Robert Lewis
              Jan 2 at 20:57












            • $begingroup$
              Yes indeed, i'm still a little confused on how you figured out a = 1 however, is this because 0 cannot exist as an eigenvector?
              $endgroup$
              – RocketKangaroo
              Jan 2 at 21:25










            • $begingroup$
              @RocketKangaroo: see (6)-(7) and the words between; they explain why $a ne 0$; then we just multiply $vec v$ by $a^{-1}$, whatever $a ne 0$ may be.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:31










            • $begingroup$
              @RocketKangaroo: yes, because $0$ cannot be an eigenvector.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:32














            3












            3








            3





            $begingroup$

            It's pretty straightforward; I'll work out the case $lambda = 3$; suppose



            $vec v = begin{pmatrix} a \ b end{pmatrix} tag 1$



            is the eigenvector corresponding to $lambda = 3$; then



            $A vec v = 3vec v; tag 2$



            we write this equation out using the given entries of $A$, $vec v$:



            $begin{bmatrix} 1 & 1 \ 4 & 1 end{bmatrix} begin{pmatrix} a \ b end{pmatrix} = 3 begin{pmatrix} a \ b end{pmatrix} = begin{pmatrix} 3a \ 3b end{pmatrix}, tag 3$



            or



            $a + b = 3a, tag 4$



            $4a + b = 3b; tag 5$



            from (4),



            $b = 2a; tag 6$



            it is expeditious at this point to observe we cannot have an eigenvector with $a = 0$, lest by (6) $b = 0$ as well, and thus



            $vec v = 0; tag 7$



            but eigenvectors are non-vanishing by definition. Thus



            $a ne 0, tag 8$



            and since eigenvectors are scalable, that is, the quality of "eigenvector-ness" is invariant under scalar multiplication, we may assume



            $a = 1; tag 9$



            then



            $b = 2a = 2, tag{10}$



            $vec v = begin{pmatrix} 1 \ 2 end{pmatrix}; tag{11}$



            it is easy to see that



            $4a + b = 4(1) + 2 = 6 = 3(2), tag{12}$



            so $vec v$ also satisfies (5).



            That's about it for the case $lambda = 3$; I leave the case $lambda = -1$ to the reader.






            share|cite|improve this answer









            $endgroup$



            It's pretty straightforward; I'll work out the case $lambda = 3$; suppose



            $vec v = begin{pmatrix} a \ b end{pmatrix} tag 1$



            is the eigenvector corresponding to $lambda = 3$; then



            $A vec v = 3vec v; tag 2$



            we write this equation out using the given entries of $A$, $vec v$:



            $begin{bmatrix} 1 & 1 \ 4 & 1 end{bmatrix} begin{pmatrix} a \ b end{pmatrix} = 3 begin{pmatrix} a \ b end{pmatrix} = begin{pmatrix} 3a \ 3b end{pmatrix}, tag 3$



            or



            $a + b = 3a, tag 4$



            $4a + b = 3b; tag 5$



            from (4),



            $b = 2a; tag 6$



            it is expeditious at this point to observe we cannot have an eigenvector with $a = 0$, lest by (6) $b = 0$ as well, and thus



            $vec v = 0; tag 7$



            but eigenvectors are non-vanishing by definition. Thus



            $a ne 0, tag 8$



            and since eigenvectors are scalable, that is, the quality of "eigenvector-ness" is invariant under scalar multiplication, we may assume



            $a = 1; tag 9$



            then



            $b = 2a = 2, tag{10}$



            $vec v = begin{pmatrix} 1 \ 2 end{pmatrix}; tag{11}$



            it is easy to see that



            $4a + b = 4(1) + 2 = 6 = 3(2), tag{12}$



            so $vec v$ also satisfies (5).



            That's about it for the case $lambda = 3$; I leave the case $lambda = -1$ to the reader.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 19:50









            Robert LewisRobert Lewis

            47.8k23067




            47.8k23067








            • 1




              $begingroup$
              Hi, could you expand on how you went from (3) to (6) please? I'm getting confused because the materials i am working from are telling me to use the following formula for a 2x2 matrix; (a-λ)x+by=0 and cx+(d-λ)y=0.
              $endgroup$
              – RocketKangaroo
              Jan 2 at 20:49












            • $begingroup$
              @RocketKangaroo: look at (4), which is just the first row of (3) written out. (4) says $a + b = 3a$; now subtract $a$ from both sides; you get (6), OK? The formula $(a - lambda)x + by = 0$ is also the first row of $Avec w = lambda vec w$ written out, with $A = begin{bmatrix} a & b \ c & d end{bmatrix}$ and $vec w = (x, y)^T$; from $A vec w = lambda vec w$ we get $ax + by = lambda a$ or $(a - lambda)x + by = 0$. It's just the general case of what I did. Does this help?
              $endgroup$
              – Robert Lewis
              Jan 2 at 20:57












            • $begingroup$
              Yes indeed, i'm still a little confused on how you figured out a = 1 however, is this because 0 cannot exist as an eigenvector?
              $endgroup$
              – RocketKangaroo
              Jan 2 at 21:25










            • $begingroup$
              @RocketKangaroo: see (6)-(7) and the words between; they explain why $a ne 0$; then we just multiply $vec v$ by $a^{-1}$, whatever $a ne 0$ may be.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:31










            • $begingroup$
              @RocketKangaroo: yes, because $0$ cannot be an eigenvector.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:32














            • 1




              $begingroup$
              Hi, could you expand on how you went from (3) to (6) please? I'm getting confused because the materials i am working from are telling me to use the following formula for a 2x2 matrix; (a-λ)x+by=0 and cx+(d-λ)y=0.
              $endgroup$
              – RocketKangaroo
              Jan 2 at 20:49












            • $begingroup$
              @RocketKangaroo: look at (4), which is just the first row of (3) written out. (4) says $a + b = 3a$; now subtract $a$ from both sides; you get (6), OK? The formula $(a - lambda)x + by = 0$ is also the first row of $Avec w = lambda vec w$ written out, with $A = begin{bmatrix} a & b \ c & d end{bmatrix}$ and $vec w = (x, y)^T$; from $A vec w = lambda vec w$ we get $ax + by = lambda a$ or $(a - lambda)x + by = 0$. It's just the general case of what I did. Does this help?
              $endgroup$
              – Robert Lewis
              Jan 2 at 20:57












            • $begingroup$
              Yes indeed, i'm still a little confused on how you figured out a = 1 however, is this because 0 cannot exist as an eigenvector?
              $endgroup$
              – RocketKangaroo
              Jan 2 at 21:25










            • $begingroup$
              @RocketKangaroo: see (6)-(7) and the words between; they explain why $a ne 0$; then we just multiply $vec v$ by $a^{-1}$, whatever $a ne 0$ may be.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:31










            • $begingroup$
              @RocketKangaroo: yes, because $0$ cannot be an eigenvector.
              $endgroup$
              – Robert Lewis
              Jan 2 at 21:32








            1




            1




            $begingroup$
            Hi, could you expand on how you went from (3) to (6) please? I'm getting confused because the materials i am working from are telling me to use the following formula for a 2x2 matrix; (a-λ)x+by=0 and cx+(d-λ)y=0.
            $endgroup$
            – RocketKangaroo
            Jan 2 at 20:49






            $begingroup$
            Hi, could you expand on how you went from (3) to (6) please? I'm getting confused because the materials i am working from are telling me to use the following formula for a 2x2 matrix; (a-λ)x+by=0 and cx+(d-λ)y=0.
            $endgroup$
            – RocketKangaroo
            Jan 2 at 20:49














            $begingroup$
            @RocketKangaroo: look at (4), which is just the first row of (3) written out. (4) says $a + b = 3a$; now subtract $a$ from both sides; you get (6), OK? The formula $(a - lambda)x + by = 0$ is also the first row of $Avec w = lambda vec w$ written out, with $A = begin{bmatrix} a & b \ c & d end{bmatrix}$ and $vec w = (x, y)^T$; from $A vec w = lambda vec w$ we get $ax + by = lambda a$ or $(a - lambda)x + by = 0$. It's just the general case of what I did. Does this help?
            $endgroup$
            – Robert Lewis
            Jan 2 at 20:57






            $begingroup$
            @RocketKangaroo: look at (4), which is just the first row of (3) written out. (4) says $a + b = 3a$; now subtract $a$ from both sides; you get (6), OK? The formula $(a - lambda)x + by = 0$ is also the first row of $Avec w = lambda vec w$ written out, with $A = begin{bmatrix} a & b \ c & d end{bmatrix}$ and $vec w = (x, y)^T$; from $A vec w = lambda vec w$ we get $ax + by = lambda a$ or $(a - lambda)x + by = 0$. It's just the general case of what I did. Does this help?
            $endgroup$
            – Robert Lewis
            Jan 2 at 20:57














            $begingroup$
            Yes indeed, i'm still a little confused on how you figured out a = 1 however, is this because 0 cannot exist as an eigenvector?
            $endgroup$
            – RocketKangaroo
            Jan 2 at 21:25




            $begingroup$
            Yes indeed, i'm still a little confused on how you figured out a = 1 however, is this because 0 cannot exist as an eigenvector?
            $endgroup$
            – RocketKangaroo
            Jan 2 at 21:25












            $begingroup$
            @RocketKangaroo: see (6)-(7) and the words between; they explain why $a ne 0$; then we just multiply $vec v$ by $a^{-1}$, whatever $a ne 0$ may be.
            $endgroup$
            – Robert Lewis
            Jan 2 at 21:31




            $begingroup$
            @RocketKangaroo: see (6)-(7) and the words between; they explain why $a ne 0$; then we just multiply $vec v$ by $a^{-1}$, whatever $a ne 0$ may be.
            $endgroup$
            – Robert Lewis
            Jan 2 at 21:31












            $begingroup$
            @RocketKangaroo: yes, because $0$ cannot be an eigenvector.
            $endgroup$
            – Robert Lewis
            Jan 2 at 21:32




            $begingroup$
            @RocketKangaroo: yes, because $0$ cannot be an eigenvector.
            $endgroup$
            – Robert Lewis
            Jan 2 at 21:32











            2












            $begingroup$

            To find an eigenvector of $lambda=3$ you need to solve :
            $$AX = lambda X $$
            $$begin{bmatrix}1&1\4&1end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}=3begin{bmatrix}x_1\x_2end{bmatrix}$$
            $$x_1+x_2 = 3x_1 text{ and }4x_1+x_2=3x_2 $$
            Thus $X=begin{bmatrix}1\2end{bmatrix}$ is an eigenvector for $lambda=3$.



            You can do the same reasonning for $lambda=-1$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Gee, your answer is almost the same as mine! How could that possible be? ;) +1
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:51






            • 1




              $begingroup$
              Yours is way more detailed :)
              $endgroup$
              – Jennifer
              Jan 2 at 19:53










            • $begingroup$
              True enough. Maybe I talk too much! ;)
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:55
















            2












            $begingroup$

            To find an eigenvector of $lambda=3$ you need to solve :
            $$AX = lambda X $$
            $$begin{bmatrix}1&1\4&1end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}=3begin{bmatrix}x_1\x_2end{bmatrix}$$
            $$x_1+x_2 = 3x_1 text{ and }4x_1+x_2=3x_2 $$
            Thus $X=begin{bmatrix}1\2end{bmatrix}$ is an eigenvector for $lambda=3$.



            You can do the same reasonning for $lambda=-1$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Gee, your answer is almost the same as mine! How could that possible be? ;) +1
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:51






            • 1




              $begingroup$
              Yours is way more detailed :)
              $endgroup$
              – Jennifer
              Jan 2 at 19:53










            • $begingroup$
              True enough. Maybe I talk too much! ;)
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:55














            2












            2








            2





            $begingroup$

            To find an eigenvector of $lambda=3$ you need to solve :
            $$AX = lambda X $$
            $$begin{bmatrix}1&1\4&1end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}=3begin{bmatrix}x_1\x_2end{bmatrix}$$
            $$x_1+x_2 = 3x_1 text{ and }4x_1+x_2=3x_2 $$
            Thus $X=begin{bmatrix}1\2end{bmatrix}$ is an eigenvector for $lambda=3$.



            You can do the same reasonning for $lambda=-1$.






            share|cite|improve this answer









            $endgroup$



            To find an eigenvector of $lambda=3$ you need to solve :
            $$AX = lambda X $$
            $$begin{bmatrix}1&1\4&1end{bmatrix}begin{bmatrix}x_1\x_2end{bmatrix}=3begin{bmatrix}x_1\x_2end{bmatrix}$$
            $$x_1+x_2 = 3x_1 text{ and }4x_1+x_2=3x_2 $$
            Thus $X=begin{bmatrix}1\2end{bmatrix}$ is an eigenvector for $lambda=3$.



            You can do the same reasonning for $lambda=-1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 19:38









            JenniferJennifer

            8,44721837




            8,44721837








            • 1




              $begingroup$
              Gee, your answer is almost the same as mine! How could that possible be? ;) +1
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:51






            • 1




              $begingroup$
              Yours is way more detailed :)
              $endgroup$
              – Jennifer
              Jan 2 at 19:53










            • $begingroup$
              True enough. Maybe I talk too much! ;)
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:55














            • 1




              $begingroup$
              Gee, your answer is almost the same as mine! How could that possible be? ;) +1
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:51






            • 1




              $begingroup$
              Yours is way more detailed :)
              $endgroup$
              – Jennifer
              Jan 2 at 19:53










            • $begingroup$
              True enough. Maybe I talk too much! ;)
              $endgroup$
              – Robert Lewis
              Jan 2 at 19:55








            1




            1




            $begingroup$
            Gee, your answer is almost the same as mine! How could that possible be? ;) +1
            $endgroup$
            – Robert Lewis
            Jan 2 at 19:51




            $begingroup$
            Gee, your answer is almost the same as mine! How could that possible be? ;) +1
            $endgroup$
            – Robert Lewis
            Jan 2 at 19:51




            1




            1




            $begingroup$
            Yours is way more detailed :)
            $endgroup$
            – Jennifer
            Jan 2 at 19:53




            $begingroup$
            Yours is way more detailed :)
            $endgroup$
            – Jennifer
            Jan 2 at 19:53












            $begingroup$
            True enough. Maybe I talk too much! ;)
            $endgroup$
            – Robert Lewis
            Jan 2 at 19:55




            $begingroup$
            True enough. Maybe I talk too much! ;)
            $endgroup$
            – Robert Lewis
            Jan 2 at 19:55











            0












            $begingroup$

            The characteristic polynomial of $A$ is
            $$
            (lambda-1)^2-4=(lambda-1-2)(lambda-1+2)=(lambda-3)(lambda+1)
            $$
            Therefore $(A-3I)(A+I)=(A+I)(A-3I)=0$.
            So the columns of $(A-3I)$ are solutions of $(A+I)X=0$. And the columns of $(A+I)$ are solutions of $(A-3I)X=0$.



            Concretely,
            $$
            A-3I = left[begin{array}{cc}-2 & 1 \ 4 & -2end{array}right]
            $$
            So
            $$
            (A+I)left[begin{array}{c} 1 \ -2 end{array}right]=0
            $$
            And
            $$
            A+I = left[begin{array}{cc} 2 & 1 \ 4 & 2end{array}right]
            $$
            So
            $$
            (A-3I)left[begin{array}{c}1 \ 2end{array}right] = 0.
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The characteristic polynomial of $A$ is
              $$
              (lambda-1)^2-4=(lambda-1-2)(lambda-1+2)=(lambda-3)(lambda+1)
              $$
              Therefore $(A-3I)(A+I)=(A+I)(A-3I)=0$.
              So the columns of $(A-3I)$ are solutions of $(A+I)X=0$. And the columns of $(A+I)$ are solutions of $(A-3I)X=0$.



              Concretely,
              $$
              A-3I = left[begin{array}{cc}-2 & 1 \ 4 & -2end{array}right]
              $$
              So
              $$
              (A+I)left[begin{array}{c} 1 \ -2 end{array}right]=0
              $$
              And
              $$
              A+I = left[begin{array}{cc} 2 & 1 \ 4 & 2end{array}right]
              $$
              So
              $$
              (A-3I)left[begin{array}{c}1 \ 2end{array}right] = 0.
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The characteristic polynomial of $A$ is
                $$
                (lambda-1)^2-4=(lambda-1-2)(lambda-1+2)=(lambda-3)(lambda+1)
                $$
                Therefore $(A-3I)(A+I)=(A+I)(A-3I)=0$.
                So the columns of $(A-3I)$ are solutions of $(A+I)X=0$. And the columns of $(A+I)$ are solutions of $(A-3I)X=0$.



                Concretely,
                $$
                A-3I = left[begin{array}{cc}-2 & 1 \ 4 & -2end{array}right]
                $$
                So
                $$
                (A+I)left[begin{array}{c} 1 \ -2 end{array}right]=0
                $$
                And
                $$
                A+I = left[begin{array}{cc} 2 & 1 \ 4 & 2end{array}right]
                $$
                So
                $$
                (A-3I)left[begin{array}{c}1 \ 2end{array}right] = 0.
                $$






                share|cite|improve this answer









                $endgroup$



                The characteristic polynomial of $A$ is
                $$
                (lambda-1)^2-4=(lambda-1-2)(lambda-1+2)=(lambda-3)(lambda+1)
                $$
                Therefore $(A-3I)(A+I)=(A+I)(A-3I)=0$.
                So the columns of $(A-3I)$ are solutions of $(A+I)X=0$. And the columns of $(A+I)$ are solutions of $(A-3I)X=0$.



                Concretely,
                $$
                A-3I = left[begin{array}{cc}-2 & 1 \ 4 & -2end{array}right]
                $$
                So
                $$
                (A+I)left[begin{array}{c} 1 \ -2 end{array}right]=0
                $$
                And
                $$
                A+I = left[begin{array}{cc} 2 & 1 \ 4 & 2end{array}right]
                $$
                So
                $$
                (A-3I)left[begin{array}{c}1 \ 2end{array}right] = 0.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 5:13









                DisintegratingByPartsDisintegratingByParts

                59.6k42581




                59.6k42581






























                    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%2f3059856%2fhow-can-i-solve-the-eigenvector-equation-for-a-2-times-2-matrix-with-eigenval%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

                    Bressuire

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more

                    Cabo Verde

                    Image
                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                    Read more

                    Gyllenstierna

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                    Read more