Intuition of generalized eigenvector.












1












$begingroup$


I was trying to get an intuitive grasp about what the the generalized eigenvector intuitively is. I read this nice answer, so I understand that in the basis given by the generalized eigenvectors, a jordan block is a linear map that is the sum of a stretch by a factor $lambda$ (eigenvalue associated to the block) and a "collapse", but I don't understand the conclusion on what these famous generalized eigenvectors actually are...




Thus the kernel of $(T−λI)k$ picks up all the Jordan blocks associated with eigenvalue $λ$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $λ$, up to some "error" term generated by certain of the other generalized eigenvectors.




Maybe someone that actually understand the last argument can care to explain with some more detail? Thank you.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not sure where the "collapse" came from. I would talk about a generalized shear. In the case of a $2times 2$ block, it is literally a stretched shear.
    $endgroup$
    – Ted Shifrin
    1 hour ago
















1












$begingroup$


I was trying to get an intuitive grasp about what the the generalized eigenvector intuitively is. I read this nice answer, so I understand that in the basis given by the generalized eigenvectors, a jordan block is a linear map that is the sum of a stretch by a factor $lambda$ (eigenvalue associated to the block) and a "collapse", but I don't understand the conclusion on what these famous generalized eigenvectors actually are...




Thus the kernel of $(T−λI)k$ picks up all the Jordan blocks associated with eigenvalue $λ$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $λ$, up to some "error" term generated by certain of the other generalized eigenvectors.




Maybe someone that actually understand the last argument can care to explain with some more detail? Thank you.










share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not sure where the "collapse" came from. I would talk about a generalized shear. In the case of a $2times 2$ block, it is literally a stretched shear.
    $endgroup$
    – Ted Shifrin
    1 hour ago














1












1








1





$begingroup$


I was trying to get an intuitive grasp about what the the generalized eigenvector intuitively is. I read this nice answer, so I understand that in the basis given by the generalized eigenvectors, a jordan block is a linear map that is the sum of a stretch by a factor $lambda$ (eigenvalue associated to the block) and a "collapse", but I don't understand the conclusion on what these famous generalized eigenvectors actually are...




Thus the kernel of $(T−λI)k$ picks up all the Jordan blocks associated with eigenvalue $λ$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $λ$, up to some "error" term generated by certain of the other generalized eigenvectors.




Maybe someone that actually understand the last argument can care to explain with some more detail? Thank you.










share|cite|improve this question











$endgroup$




I was trying to get an intuitive grasp about what the the generalized eigenvector intuitively is. I read this nice answer, so I understand that in the basis given by the generalized eigenvectors, a jordan block is a linear map that is the sum of a stretch by a factor $lambda$ (eigenvalue associated to the block) and a "collapse", but I don't understand the conclusion on what these famous generalized eigenvectors actually are...




Thus the kernel of $(T−λI)k$ picks up all the Jordan blocks associated with eigenvalue $λ$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $λ$, up to some "error" term generated by certain of the other generalized eigenvectors.




Maybe someone that actually understand the last argument can care to explain with some more detail? Thank you.







linear-algebra intuition jordan-normal-form generalizedeigenvector






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Andrews

1,2761421




1,2761421










asked 2 hours ago









roi_saumonroi_saumon

62338




62338












  • $begingroup$
    I'm not sure where the "collapse" came from. I would talk about a generalized shear. In the case of a $2times 2$ block, it is literally a stretched shear.
    $endgroup$
    – Ted Shifrin
    1 hour ago


















  • $begingroup$
    I'm not sure where the "collapse" came from. I would talk about a generalized shear. In the case of a $2times 2$ block, it is literally a stretched shear.
    $endgroup$
    – Ted Shifrin
    1 hour ago
















$begingroup$
I'm not sure where the "collapse" came from. I would talk about a generalized shear. In the case of a $2times 2$ block, it is literally a stretched shear.
$endgroup$
– Ted Shifrin
1 hour ago




$begingroup$
I'm not sure where the "collapse" came from. I would talk about a generalized shear. In the case of a $2times 2$ block, it is literally a stretched shear.
$endgroup$
– Ted Shifrin
1 hour ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Don't look for anything particularly deep or fancy here.



If you have a calculation to do that involves some vectors and a linear operator $T$ that you apply perhaps to several of the vectors or more several times in sequence, then it can simplify the calculation if you represent the vectors in an eigenbasis -- since then we have
$$ T(x_1,x_2,ldots,x_n) = (lambda_1x_1, lambda_2x_2, ldots, lambda_n x_n) $$
Each component of the vector just gets multiplied by its associated eigenvalue and the different components don't interact with each other at all.



Unfortunately, not all operators can be written in this nice form, because there may not be enough eigenvectors to combine into a basis. In that case the "next best thing" we can do is choosing a basis where the matrix of $T$ is in Jordan form. Then each component of $T(x_1,x_2,ldots,x_n)$ is either $lambda_i x_i$ or $lambda_i x_i + x_{i+1}$, which is still somewhat simpler than multiplication by an arbitrary matrix.



Since this gives us some of what a basis consisting entirely of eigenvectors gives us, in terms of computational simplicity, if seems reasonable to describe them as a generalization of eigenvectors. Especially since in the case where we do have enough eigenvectors for an eigenbasis, generalized eigenvectors are the same as eigenvectors.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    I see generalized eigenvectors as an attempt to patch up the discrepancy between geometric multipicity and algebraic multiplicity of eigenvalues. This discrepancy is most easily observed as a result of skew transformations (and looking at Jordan normal forms, we see that skew transformations are in some sense at the core of any such discrepancy).



    For instance, take the skew transformation given by the matrix
    $$
    begin{bmatrix}1&1\0&1end{bmatrix}
    $$

    It has eigenvalue $1$ with algebraic multiplicity $2$ (the characteristic polynomial is $(lambda - 1)^2$, which has a double root at $1$), but geometric multiplicity $1$ (the eigenspace has dimension $1$, as it just the $x$-axis).



    However, the space of generalized eigenvectors is the entire plane, which is 2-dimensional, and more in line with the algebraic multiplicity of the eigenvalue.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Consider the matrix $$A= begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}$$. Obviously 3 is the only eigenvalue with "algebraic multiplicity" 2. It is also easy to find the eigenvectors $$begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}begin{bmatrix}x\ yend{bmatrix}= begin{bmatrix}3x+ y \ 3yend{bmatrix}= begin{bmatrix}3x \ 3y end{bmatrix}$$ so that we have 3x+ y= 3x and 3y= 3y. The first equation gives y= 0 and the second is satisfied for any x. So any eigenvector is of the form $$begin{bmatrix}x \ 0 end{bmatrix}= xbegin{bmatrix}1 \ 0 end{bmatrix}$$. So the subspace of all eigenvectors has dimension 1 (the geometric multiplicity is 1).



      $$v= begin{bmatrix}0 \ 1 end{bmatrix}$$ is NOT an eigenvector: $$(A- 3I)v= begin{bmatrix}0 & 1 \ 0 & 0 end{bmatrix}begin{bmatrix} 0 \ 1 end{bmatrix}= begin{bmatrix}1 \ 0end{bmatrix}$$, NOT the zero vector. But it does give the previous eigenvector so that applying A- 3I again would give the 0 vector it is a "generalized eigenvector.






      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%2f3159938%2fintuition-of-generalized-eigenvector%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









        2












        $begingroup$

        Don't look for anything particularly deep or fancy here.



        If you have a calculation to do that involves some vectors and a linear operator $T$ that you apply perhaps to several of the vectors or more several times in sequence, then it can simplify the calculation if you represent the vectors in an eigenbasis -- since then we have
        $$ T(x_1,x_2,ldots,x_n) = (lambda_1x_1, lambda_2x_2, ldots, lambda_n x_n) $$
        Each component of the vector just gets multiplied by its associated eigenvalue and the different components don't interact with each other at all.



        Unfortunately, not all operators can be written in this nice form, because there may not be enough eigenvectors to combine into a basis. In that case the "next best thing" we can do is choosing a basis where the matrix of $T$ is in Jordan form. Then each component of $T(x_1,x_2,ldots,x_n)$ is either $lambda_i x_i$ or $lambda_i x_i + x_{i+1}$, which is still somewhat simpler than multiplication by an arbitrary matrix.



        Since this gives us some of what a basis consisting entirely of eigenvectors gives us, in terms of computational simplicity, if seems reasonable to describe them as a generalization of eigenvectors. Especially since in the case where we do have enough eigenvectors for an eigenbasis, generalized eigenvectors are the same as eigenvectors.






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          Don't look for anything particularly deep or fancy here.



          If you have a calculation to do that involves some vectors and a linear operator $T$ that you apply perhaps to several of the vectors or more several times in sequence, then it can simplify the calculation if you represent the vectors in an eigenbasis -- since then we have
          $$ T(x_1,x_2,ldots,x_n) = (lambda_1x_1, lambda_2x_2, ldots, lambda_n x_n) $$
          Each component of the vector just gets multiplied by its associated eigenvalue and the different components don't interact with each other at all.



          Unfortunately, not all operators can be written in this nice form, because there may not be enough eigenvectors to combine into a basis. In that case the "next best thing" we can do is choosing a basis where the matrix of $T$ is in Jordan form. Then each component of $T(x_1,x_2,ldots,x_n)$ is either $lambda_i x_i$ or $lambda_i x_i + x_{i+1}$, which is still somewhat simpler than multiplication by an arbitrary matrix.



          Since this gives us some of what a basis consisting entirely of eigenvectors gives us, in terms of computational simplicity, if seems reasonable to describe them as a generalization of eigenvectors. Especially since in the case where we do have enough eigenvectors for an eigenbasis, generalized eigenvectors are the same as eigenvectors.






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            Don't look for anything particularly deep or fancy here.



            If you have a calculation to do that involves some vectors and a linear operator $T$ that you apply perhaps to several of the vectors or more several times in sequence, then it can simplify the calculation if you represent the vectors in an eigenbasis -- since then we have
            $$ T(x_1,x_2,ldots,x_n) = (lambda_1x_1, lambda_2x_2, ldots, lambda_n x_n) $$
            Each component of the vector just gets multiplied by its associated eigenvalue and the different components don't interact with each other at all.



            Unfortunately, not all operators can be written in this nice form, because there may not be enough eigenvectors to combine into a basis. In that case the "next best thing" we can do is choosing a basis where the matrix of $T$ is in Jordan form. Then each component of $T(x_1,x_2,ldots,x_n)$ is either $lambda_i x_i$ or $lambda_i x_i + x_{i+1}$, which is still somewhat simpler than multiplication by an arbitrary matrix.



            Since this gives us some of what a basis consisting entirely of eigenvectors gives us, in terms of computational simplicity, if seems reasonable to describe them as a generalization of eigenvectors. Especially since in the case where we do have enough eigenvectors for an eigenbasis, generalized eigenvectors are the same as eigenvectors.






            share|cite|improve this answer











            $endgroup$



            Don't look for anything particularly deep or fancy here.



            If you have a calculation to do that involves some vectors and a linear operator $T$ that you apply perhaps to several of the vectors or more several times in sequence, then it can simplify the calculation if you represent the vectors in an eigenbasis -- since then we have
            $$ T(x_1,x_2,ldots,x_n) = (lambda_1x_1, lambda_2x_2, ldots, lambda_n x_n) $$
            Each component of the vector just gets multiplied by its associated eigenvalue and the different components don't interact with each other at all.



            Unfortunately, not all operators can be written in this nice form, because there may not be enough eigenvectors to combine into a basis. In that case the "next best thing" we can do is choosing a basis where the matrix of $T$ is in Jordan form. Then each component of $T(x_1,x_2,ldots,x_n)$ is either $lambda_i x_i$ or $lambda_i x_i + x_{i+1}$, which is still somewhat simpler than multiplication by an arbitrary matrix.



            Since this gives us some of what a basis consisting entirely of eigenvectors gives us, in terms of computational simplicity, if seems reasonable to describe them as a generalization of eigenvectors. Especially since in the case where we do have enough eigenvectors for an eigenbasis, generalized eigenvectors are the same as eigenvectors.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            Henning MakholmHenning Makholm

            242k17308550




            242k17308550























                1












                $begingroup$

                I see generalized eigenvectors as an attempt to patch up the discrepancy between geometric multipicity and algebraic multiplicity of eigenvalues. This discrepancy is most easily observed as a result of skew transformations (and looking at Jordan normal forms, we see that skew transformations are in some sense at the core of any such discrepancy).



                For instance, take the skew transformation given by the matrix
                $$
                begin{bmatrix}1&1\0&1end{bmatrix}
                $$

                It has eigenvalue $1$ with algebraic multiplicity $2$ (the characteristic polynomial is $(lambda - 1)^2$, which has a double root at $1$), but geometric multiplicity $1$ (the eigenspace has dimension $1$, as it just the $x$-axis).



                However, the space of generalized eigenvectors is the entire plane, which is 2-dimensional, and more in line with the algebraic multiplicity of the eigenvalue.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  I see generalized eigenvectors as an attempt to patch up the discrepancy between geometric multipicity and algebraic multiplicity of eigenvalues. This discrepancy is most easily observed as a result of skew transformations (and looking at Jordan normal forms, we see that skew transformations are in some sense at the core of any such discrepancy).



                  For instance, take the skew transformation given by the matrix
                  $$
                  begin{bmatrix}1&1\0&1end{bmatrix}
                  $$

                  It has eigenvalue $1$ with algebraic multiplicity $2$ (the characteristic polynomial is $(lambda - 1)^2$, which has a double root at $1$), but geometric multiplicity $1$ (the eigenspace has dimension $1$, as it just the $x$-axis).



                  However, the space of generalized eigenvectors is the entire plane, which is 2-dimensional, and more in line with the algebraic multiplicity of the eigenvalue.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    I see generalized eigenvectors as an attempt to patch up the discrepancy between geometric multipicity and algebraic multiplicity of eigenvalues. This discrepancy is most easily observed as a result of skew transformations (and looking at Jordan normal forms, we see that skew transformations are in some sense at the core of any such discrepancy).



                    For instance, take the skew transformation given by the matrix
                    $$
                    begin{bmatrix}1&1\0&1end{bmatrix}
                    $$

                    It has eigenvalue $1$ with algebraic multiplicity $2$ (the characteristic polynomial is $(lambda - 1)^2$, which has a double root at $1$), but geometric multiplicity $1$ (the eigenspace has dimension $1$, as it just the $x$-axis).



                    However, the space of generalized eigenvectors is the entire plane, which is 2-dimensional, and more in line with the algebraic multiplicity of the eigenvalue.






                    share|cite|improve this answer









                    $endgroup$



                    I see generalized eigenvectors as an attempt to patch up the discrepancy between geometric multipicity and algebraic multiplicity of eigenvalues. This discrepancy is most easily observed as a result of skew transformations (and looking at Jordan normal forms, we see that skew transformations are in some sense at the core of any such discrepancy).



                    For instance, take the skew transformation given by the matrix
                    $$
                    begin{bmatrix}1&1\0&1end{bmatrix}
                    $$

                    It has eigenvalue $1$ with algebraic multiplicity $2$ (the characteristic polynomial is $(lambda - 1)^2$, which has a double root at $1$), but geometric multiplicity $1$ (the eigenspace has dimension $1$, as it just the $x$-axis).



                    However, the space of generalized eigenvectors is the entire plane, which is 2-dimensional, and more in line with the algebraic multiplicity of the eigenvalue.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    ArthurArthur

                    119k7118202




                    119k7118202























                        0












                        $begingroup$

                        Consider the matrix $$A= begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}$$. Obviously 3 is the only eigenvalue with "algebraic multiplicity" 2. It is also easy to find the eigenvectors $$begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}begin{bmatrix}x\ yend{bmatrix}= begin{bmatrix}3x+ y \ 3yend{bmatrix}= begin{bmatrix}3x \ 3y end{bmatrix}$$ so that we have 3x+ y= 3x and 3y= 3y. The first equation gives y= 0 and the second is satisfied for any x. So any eigenvector is of the form $$begin{bmatrix}x \ 0 end{bmatrix}= xbegin{bmatrix}1 \ 0 end{bmatrix}$$. So the subspace of all eigenvectors has dimension 1 (the geometric multiplicity is 1).



                        $$v= begin{bmatrix}0 \ 1 end{bmatrix}$$ is NOT an eigenvector: $$(A- 3I)v= begin{bmatrix}0 & 1 \ 0 & 0 end{bmatrix}begin{bmatrix} 0 \ 1 end{bmatrix}= begin{bmatrix}1 \ 0end{bmatrix}$$, NOT the zero vector. But it does give the previous eigenvector so that applying A- 3I again would give the 0 vector it is a "generalized eigenvector.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Consider the matrix $$A= begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}$$. Obviously 3 is the only eigenvalue with "algebraic multiplicity" 2. It is also easy to find the eigenvectors $$begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}begin{bmatrix}x\ yend{bmatrix}= begin{bmatrix}3x+ y \ 3yend{bmatrix}= begin{bmatrix}3x \ 3y end{bmatrix}$$ so that we have 3x+ y= 3x and 3y= 3y. The first equation gives y= 0 and the second is satisfied for any x. So any eigenvector is of the form $$begin{bmatrix}x \ 0 end{bmatrix}= xbegin{bmatrix}1 \ 0 end{bmatrix}$$. So the subspace of all eigenvectors has dimension 1 (the geometric multiplicity is 1).



                          $$v= begin{bmatrix}0 \ 1 end{bmatrix}$$ is NOT an eigenvector: $$(A- 3I)v= begin{bmatrix}0 & 1 \ 0 & 0 end{bmatrix}begin{bmatrix} 0 \ 1 end{bmatrix}= begin{bmatrix}1 \ 0end{bmatrix}$$, NOT the zero vector. But it does give the previous eigenvector so that applying A- 3I again would give the 0 vector it is a "generalized eigenvector.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Consider the matrix $$A= begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}$$. Obviously 3 is the only eigenvalue with "algebraic multiplicity" 2. It is also easy to find the eigenvectors $$begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}begin{bmatrix}x\ yend{bmatrix}= begin{bmatrix}3x+ y \ 3yend{bmatrix}= begin{bmatrix}3x \ 3y end{bmatrix}$$ so that we have 3x+ y= 3x and 3y= 3y. The first equation gives y= 0 and the second is satisfied for any x. So any eigenvector is of the form $$begin{bmatrix}x \ 0 end{bmatrix}= xbegin{bmatrix}1 \ 0 end{bmatrix}$$. So the subspace of all eigenvectors has dimension 1 (the geometric multiplicity is 1).



                            $$v= begin{bmatrix}0 \ 1 end{bmatrix}$$ is NOT an eigenvector: $$(A- 3I)v= begin{bmatrix}0 & 1 \ 0 & 0 end{bmatrix}begin{bmatrix} 0 \ 1 end{bmatrix}= begin{bmatrix}1 \ 0end{bmatrix}$$, NOT the zero vector. But it does give the previous eigenvector so that applying A- 3I again would give the 0 vector it is a "generalized eigenvector.






                            share|cite|improve this answer









                            $endgroup$



                            Consider the matrix $$A= begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}$$. Obviously 3 is the only eigenvalue with "algebraic multiplicity" 2. It is also easy to find the eigenvectors $$begin{bmatrix}3 & 1 \ 0 & 3 end{bmatrix}begin{bmatrix}x\ yend{bmatrix}= begin{bmatrix}3x+ y \ 3yend{bmatrix}= begin{bmatrix}3x \ 3y end{bmatrix}$$ so that we have 3x+ y= 3x and 3y= 3y. The first equation gives y= 0 and the second is satisfied for any x. So any eigenvector is of the form $$begin{bmatrix}x \ 0 end{bmatrix}= xbegin{bmatrix}1 \ 0 end{bmatrix}$$. So the subspace of all eigenvectors has dimension 1 (the geometric multiplicity is 1).



                            $$v= begin{bmatrix}0 \ 1 end{bmatrix}$$ is NOT an eigenvector: $$(A- 3I)v= begin{bmatrix}0 & 1 \ 0 & 0 end{bmatrix}begin{bmatrix} 0 \ 1 end{bmatrix}= begin{bmatrix}1 \ 0end{bmatrix}$$, NOT the zero vector. But it does give the previous eigenvector so that applying A- 3I again would give the 0 vector it is a "generalized eigenvector.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            user247327user247327

                            11.5k1516




                            11.5k1516






























                                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%2f3159938%2fintuition-of-generalized-eigenvector%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Сан-Квентин

                                Алькесар

                                Josef Freinademetz