Why does subtracting an eigenvalue from the diagonal and solving give the eigenspace?












0












$begingroup$


Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?



Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.



I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.






linear-algebra matrices eigenvalues-eigenvectors







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$endgroup$












  • $begingroup$
    Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:16












  • $begingroup$
    Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
    $endgroup$
    – Ian
    Dec 20 '18 at 10:19












  • $begingroup$
    Thanks, that makes perfect sense.
    $endgroup$
    – Eoin Ó Coinnigh
    Dec 20 '18 at 13:23
















0












$begingroup$


Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?



Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.



I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:16












  • $begingroup$
    Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
    $endgroup$
    – Ian
    Dec 20 '18 at 10:19












  • $begingroup$
    Thanks, that makes perfect sense.
    $endgroup$
    – Eoin Ó Coinnigh
    Dec 20 '18 at 13:23














0












0








0





$begingroup$


Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?



Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.



I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




Given a matrix $A$, with eigenvalue $lambda$, why does substracting $lambda I$ from $A$, and then solving the matrix give us the eigenspace?



Like, I don't understand the intuition behind subtracting the two, to hence obtain the eigenspace.



I can't find an answer anywhere, a high level explanation would be very much appreciated! Thank you in advance.






linear-algebra matrices eigenvalues-eigenvectors




linear-algebra matrices eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 10:18







Eoin Ó Coinnigh

















asked Dec 20 '18 at 10:11









Eoin Ó CoinnighEoin Ó Coinnigh

11




11












  • $begingroup$
    Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:16












  • $begingroup$
    Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
    $endgroup$
    – Ian
    Dec 20 '18 at 10:19












  • $begingroup$
    Thanks, that makes perfect sense.
    $endgroup$
    – Eoin Ó Coinnigh
    Dec 20 '18 at 13:23


















  • $begingroup$
    Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 10:16












  • $begingroup$
    Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
    $endgroup$
    – Ian
    Dec 20 '18 at 10:19












  • $begingroup$
    Thanks, that makes perfect sense.
    $endgroup$
    – Eoin Ó Coinnigh
    Dec 20 '18 at 13:23
















$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16






$begingroup$
Eigenvector by definition is the non-zero $vin V$ such that $Av=lambda v$ where $lambda $ is the corressponding eigenvalue. So to find $v$: $Av=lambda viff (A-lambda I)v=0$, where $0$ is the zero matrix. In other words eigenspace associated to $lambda$ is the $N(A-lambda I)$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:16














$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19






$begingroup$
Eigenvectors are nonzero vectors that $A-lambda I$ sends to the zero vector. This can only happen if $lambda$ is chosen so that $A-lambda I$ has zero determinant (otherwise it would be invertible so it would only send the zero vector to the zero vector). That said, finding roots of the characteristic polynomial only gets you the eigenvalues, not the eigenvectors.
$endgroup$
– Ian
Dec 20 '18 at 10:19














$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23




$begingroup$
Thanks, that makes perfect sense.
$endgroup$
– Eoin Ó Coinnigh
Dec 20 '18 at 13:23










3 Answers
3






active

oldest

votes


















1












$begingroup$

Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$



Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
$$iff vin ker(A-lambda I).$$



We conclude that $$E_lambda =ker(A-lambda I).$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
    $$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.



    On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Notice that
      begin{align*}
      Av= lambda v,~v ne 0
      Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
      end{align*}






      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









        1












        $begingroup$

        Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
        In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$



        Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
        $$iff vin ker(A-lambda I).$$



        We conclude that $$E_lambda =ker(A-lambda I).$$






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
          In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$



          Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
          $$iff vin ker(A-lambda I).$$



          We conclude that $$E_lambda =ker(A-lambda I).$$






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
            In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$



            Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
            $$iff vin ker(A-lambda I).$$



            We conclude that $$E_lambda =ker(A-lambda I).$$






            share|cite|improve this answer











            $endgroup$



            Let $V$ a vector space. The eigenspace associated to the eigenvalue $lambda $ is the set $E_lambda $ of vector $vin V$ s.t. $$Av=lambda v.$$
            In otherword, $$E_lambda ={vin Vmid Av=lambda v}.$$



            Therefore $$vin E_lambda iff Av=lambda viff (A-lambda I)v=Av-lambda v=0$$
            $$iff vin ker(A-lambda I).$$



            We conclude that $$E_lambda =ker(A-lambda I).$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 20 '18 at 20:08









            Henning Makholm

            240k17306544




            240k17306544










            answered Dec 20 '18 at 10:17









            SurbSurb

            38.1k94375




            38.1k94375























                0












                $begingroup$

                Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
                $$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.



                On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
                  $$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.



                  On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
                    $$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.



                    On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.






                    share|cite|improve this answer









                    $endgroup$



                    Let $V$ be a $n$-dimensional vector space over any field and $A$ any $(n times n)$-Matrix acting on that space. Remember the eigenvalue equation $$Av = lambda v$$ for $v neq 0$. Now, rewrite $v=I v$, where $I$ is the identity matrix, and substract it from the eigenvalue equation, hence
                    $$(A- lambda I )v = 0.$$ Let us assume for a moment that the matrix $A-lambda I$ is invertible. For that we'd multiply both sides with the inverse and we'd get $v=0$. A contradiction to our requirement that $v neq 0$. Therefore, $A-lambda I$ cannot be invertible and so $det(A-lambda I) = 0$. But the characteristic polyomial is exactly the determinant of $A-lambda I$.



                    On the other hand, if $lambda$ is a zero of the characteristic polynomial, i.e. $A-lambda I$ is not invertible, $(A-lambda I) v =0$ has a non-trivial solution and its solution is an eigenvector corresponding to the eigenvalue $lambda$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 20 '18 at 10:28









                    YoungMathYoungMath

                    190110




                    190110























                        0












                        $begingroup$

                        Notice that
                        begin{align*}
                        Av= lambda v,~v ne 0
                        Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
                        end{align*}






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Notice that
                          begin{align*}
                          Av= lambda v,~v ne 0
                          Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
                          end{align*}






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Notice that
                            begin{align*}
                            Av= lambda v,~v ne 0
                            Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
                            end{align*}






                            share|cite|improve this answer









                            $endgroup$



                            Notice that
                            begin{align*}
                            Av= lambda v,~v ne 0
                            Longleftrightarrow (A - lambda I)v = 0,~v ne 0 Longleftrightarrow v intext{Nul}(A- lambda I),~v ne 0.
                            end{align*}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 20 '18 at 20:05









                            Pietro PaparellaPietro Paparella

                            1,394615




                            1,394615






























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