Intuition of generalized eigenvector.The intuition behind generalized eigenvectorsWhy is the eigenvector of a...

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Intuition of generalized eigenvector.


The intuition behind generalized eigenvectorsWhy is the eigenvector of a covariance matrix equal to a principal component?An intuitive approach to the Jordan Normal formIntuitive meaning of right and left eigenvectorConsider a linear operator $L$ and some polynomial of it, $L'=p(L)$. Show that the minimal polynomial of $L'$ has smaller degree than that of $L$.Why are the eigenvalues of a covariance matrix equal to the variance of its eigenvectors?How to determine the length of the Jordan chain associated to an eigenvector?Generalized eigenvectors for Jordan canonical form (and theory)Finding generalized eigenvectors from a Jordan formYet Another Question Regarding Jordan FormFinding ch. polynomial and Jordan normal form of $f$ knowing $dimker f=2$ and there are $a,b$ not in $ker f$ such that $f^2(a)=0, f(b)=b$













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









      $endgroup$













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        3 Answers
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        3 Answers
        3






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        active

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        active

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        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






























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