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









          $endgroup$













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            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






























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