Finding eigenvalue based on vector space and linear operator?

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So given that $L$ is a linear operator from a vector space $V$ to itself, with $V=mathbb R_2[t]$ and $Lf = f'' - f$ I need to show that $lambda = -1$ is the only eigenvalue.



I know how to do this with standard matrices but it has been a couple of years since I've taken differential equations so I'm very rusty and unsure of how to convert this information to a matrix I can use the standard $det(A - lambda I)$ equation on. Any tips?






linear-algebra differential-equations







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    up vote
    2
    down vote

    favorite
    1












    So given that $L$ is a linear operator from a vector space $V$ to itself, with $V=mathbb R_2[t]$ and $Lf = f'' - f$ I need to show that $lambda = -1$ is the only eigenvalue.



    I know how to do this with standard matrices but it has been a couple of years since I've taken differential equations so I'm very rusty and unsure of how to convert this information to a matrix I can use the standard $det(A - lambda I)$ equation on. Any tips?






    linear-algebra differential-equations







    share|cite|improve this question























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      So given that $L$ is a linear operator from a vector space $V$ to itself, with $V=mathbb R_2[t]$ and $Lf = f'' - f$ I need to show that $lambda = -1$ is the only eigenvalue.



      I know how to do this with standard matrices but it has been a couple of years since I've taken differential equations so I'm very rusty and unsure of how to convert this information to a matrix I can use the standard $det(A - lambda I)$ equation on. Any tips?






      linear-algebra differential-equations







      share|cite|improve this question













      So given that $L$ is a linear operator from a vector space $V$ to itself, with $V=mathbb R_2[t]$ and $Lf = f'' - f$ I need to show that $lambda = -1$ is the only eigenvalue.



      I know how to do this with standard matrices but it has been a couple of years since I've taken differential equations so I'm very rusty and unsure of how to convert this information to a matrix I can use the standard $det(A - lambda I)$ equation on. Any tips?






      linear-algebra differential-equations




      linear-algebra differential-equations






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      asked 2 hours ago









      user3491700

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          2 Answers
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          up vote
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          Basis of $V$ is $1,t,t^2$



          Write the matrix with respect to this basis



          $L(1)=0-1=-1cdot1+0cdot t+0cdot t^2$



          $L(t)=0-t=0cdot 1-1cdot t+0cdot t^2$



          $L(t^2)=2-t^2=2cdot 1+0cdot t-1cdot t^2$



          So matrix is $$beginbmatrix-1 & 0&2\ 0&-1&0\0&0&-1endbmatrix$$



          Now you can clearly see $lambda =-1$ is the only eigen value!






          share|cite|improve this answer






















          • For what it's worth: cdot produces the dot $cdot$
            – Omnomnomnom
            2 hours ago










          • I will edit the post. Thanks
            – StammeringMathematician
            2 hours ago

















          up vote
          4
          down vote













          The easiest approach, I think, is to note that
          $$
          [L + operatornameid]^2f = 0
          $$
          for all polynomials $f in Bbb R_2[t]$. So, if $f$ is an eigenvector associated with eigenvalue $lambda$, then $[L + operatornameid]^2f = (lambda + 1)^2f = 0$






          share|cite|improve this answer






















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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Basis of $V$ is $1,t,t^2$



            Write the matrix with respect to this basis



            $L(1)=0-1=-1cdot1+0cdot t+0cdot t^2$



            $L(t)=0-t=0cdot 1-1cdot t+0cdot t^2$



            $L(t^2)=2-t^2=2cdot 1+0cdot t-1cdot t^2$



            So matrix is $$beginbmatrix-1 & 0&2\ 0&-1&0\0&0&-1endbmatrix$$



            Now you can clearly see $lambda =-1$ is the only eigen value!






            share|cite|improve this answer






















            • For what it's worth: cdot produces the dot $cdot$
              – Omnomnomnom
              2 hours ago










            • I will edit the post. Thanks
              – StammeringMathematician
              2 hours ago














            up vote
            2
            down vote



            accepted










            Basis of $V$ is $1,t,t^2$



            Write the matrix with respect to this basis



            $L(1)=0-1=-1cdot1+0cdot t+0cdot t^2$



            $L(t)=0-t=0cdot 1-1cdot t+0cdot t^2$



            $L(t^2)=2-t^2=2cdot 1+0cdot t-1cdot t^2$



            So matrix is $$beginbmatrix-1 & 0&2\ 0&-1&0\0&0&-1endbmatrix$$



            Now you can clearly see $lambda =-1$ is the only eigen value!






            share|cite|improve this answer






















            • For what it's worth: cdot produces the dot $cdot$
              – Omnomnomnom
              2 hours ago










            • I will edit the post. Thanks
              – StammeringMathematician
              2 hours ago












            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Basis of $V$ is $1,t,t^2$



            Write the matrix with respect to this basis



            $L(1)=0-1=-1cdot1+0cdot t+0cdot t^2$



            $L(t)=0-t=0cdot 1-1cdot t+0cdot t^2$



            $L(t^2)=2-t^2=2cdot 1+0cdot t-1cdot t^2$



            So matrix is $$beginbmatrix-1 & 0&2\ 0&-1&0\0&0&-1endbmatrix$$



            Now you can clearly see $lambda =-1$ is the only eigen value!






            share|cite|improve this answer














            Basis of $V$ is $1,t,t^2$



            Write the matrix with respect to this basis



            $L(1)=0-1=-1cdot1+0cdot t+0cdot t^2$



            $L(t)=0-t=0cdot 1-1cdot t+0cdot t^2$



            $L(t^2)=2-t^2=2cdot 1+0cdot t-1cdot t^2$



            So matrix is $$beginbmatrix-1 & 0&2\ 0&-1&0\0&0&-1endbmatrix$$



            Now you can clearly see $lambda =-1$ is the only eigen value!







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours ago









            StammeringMathematician

            1,091217




            1,091217











            • For what it's worth: cdot produces the dot $cdot$
              – Omnomnomnom
              2 hours ago










            • I will edit the post. Thanks
              – StammeringMathematician
              2 hours ago
















            • For what it's worth: cdot produces the dot $cdot$
              – Omnomnomnom
              2 hours ago










            • I will edit the post. Thanks
              – StammeringMathematician
              2 hours ago















            For what it's worth: cdot produces the dot $cdot$
            – Omnomnomnom
            2 hours ago




            For what it's worth: cdot produces the dot $cdot$
            – Omnomnomnom
            2 hours ago












            I will edit the post. Thanks
            – StammeringMathematician
            2 hours ago




            I will edit the post. Thanks
            – StammeringMathematician
            2 hours ago










            up vote
            4
            down vote













            The easiest approach, I think, is to note that
            $$
            [L + operatornameid]^2f = 0
            $$
            for all polynomials $f in Bbb R_2[t]$. So, if $f$ is an eigenvector associated with eigenvalue $lambda$, then $[L + operatornameid]^2f = (lambda + 1)^2f = 0$






            share|cite|improve this answer


























              up vote
              4
              down vote













              The easiest approach, I think, is to note that
              $$
              [L + operatornameid]^2f = 0
              $$
              for all polynomials $f in Bbb R_2[t]$. So, if $f$ is an eigenvector associated with eigenvalue $lambda$, then $[L + operatornameid]^2f = (lambda + 1)^2f = 0$






              share|cite|improve this answer
























                up vote
                4
                down vote










                up vote
                4
                down vote









                The easiest approach, I think, is to note that
                $$
                [L + operatornameid]^2f = 0
                $$
                for all polynomials $f in Bbb R_2[t]$. So, if $f$ is an eigenvector associated with eigenvalue $lambda$, then $[L + operatornameid]^2f = (lambda + 1)^2f = 0$






                share|cite|improve this answer














                The easiest approach, I think, is to note that
                $$
                [L + operatornameid]^2f = 0
                $$
                for all polynomials $f in Bbb R_2[t]$. So, if $f$ is an eigenvector associated with eigenvalue $lambda$, then $[L + operatornameid]^2f = (lambda + 1)^2f = 0$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 hours ago

























                answered 2 hours ago









                Omnomnomnom

                123k785172




                123k785172



























                     

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