Why is $(T-lambda I)^p-1(x)$ an eigenvector?

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Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be a scalar. A nonzero vector $x$ in $V$ is called a generalized eigenvector of $T$ corresponding to $lambda$ if $(T-lambda I)^p(x)=0$ for some positive integer $p$.



Notice that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$, and $p$ is the smallest positive integer for which $(T-lambda I)^p(x)=0$, then $(T-lambda I)^p-1(x)$ is an eigenvector of T corresponding to $lambda$. Therefore $lambda$ is an eigenvalue of $T$.




Are we saying that the smallest positive integer $p$ is $1$, and so $p-1=0$, and then $(T-lambda I)^p-1(x)=(T-lambda I)^0(x)=x neq 0$?




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    Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be a scalar. A nonzero vector $x$ in $V$ is called a generalized eigenvector of $T$ corresponding to $lambda$ if $(T-lambda I)^p(x)=0$ for some positive integer $p$.



    Notice that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$, and $p$ is the smallest positive integer for which $(T-lambda I)^p(x)=0$, then $(T-lambda I)^p-1(x)$ is an eigenvector of T corresponding to $lambda$. Therefore $lambda$ is an eigenvalue of $T$.




    Are we saying that the smallest positive integer $p$ is $1$, and so $p-1=0$, and then $(T-lambda I)^p-1(x)=(T-lambda I)^0(x)=x neq 0$?




    linear-algebra




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      Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be a scalar. A nonzero vector $x$ in $V$ is called a generalized eigenvector of $T$ corresponding to $lambda$ if $(T-lambda I)^p(x)=0$ for some positive integer $p$.



      Notice that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$, and $p$ is the smallest positive integer for which $(T-lambda I)^p(x)=0$, then $(T-lambda I)^p-1(x)$ is an eigenvector of T corresponding to $lambda$. Therefore $lambda$ is an eigenvalue of $T$.




      Are we saying that the smallest positive integer $p$ is $1$, and so $p-1=0$, and then $(T-lambda I)^p-1(x)=(T-lambda I)^0(x)=x neq 0$?




      linear-algebra




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      Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be a scalar. A nonzero vector $x$ in $V$ is called a generalized eigenvector of $T$ corresponding to $lambda$ if $(T-lambda I)^p(x)=0$ for some positive integer $p$.



      Notice that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$, and $p$ is the smallest positive integer for which $(T-lambda I)^p(x)=0$, then $(T-lambda I)^p-1(x)$ is an eigenvector of T corresponding to $lambda$. Therefore $lambda$ is an eigenvalue of $T$.




      Are we saying that the smallest positive integer $p$ is $1$, and so $p-1=0$, and then $(T-lambda I)^p-1(x)=(T-lambda I)^0(x)=x neq 0$?





      linear-algebra





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      edited 2 days ago









      Brahadeesh

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          We're saying that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$ then for some $n$ we have $(T-lambda I)^n(x) =0$.



          Note that this $n$ is not unique, because $(T-lambda I)^n+1(x) =0$ too. But we can choose $p$ to be the smallest such $n$ (because any non-empty set of natural numbers has a minimum element).



          Then if we set $v = (T-lambda I)^p-1(x)$ then $vneq 0$ by minimality of $p$, and we have $(T-lambda I)(v) = 0$, so $v$ is an eigenvector of $T$ corresponding to $lambda$.






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            You can see this property by expanding the definition of exponentials of linear maps. This is nothing else than repeated composition.



            You have $(T-lambda I)^p(x)=(T-lambda I)((T-lambda I)^p-1(x))=mathbf 0$, i.e. per definition $(T-lambda I)^p-1(x)inmathrmker(T-lambda I)$, i.e. $(T-lambda I)^p-1(x)$ is an eigenvector.



            Now, this eigenvector really is proper(i.e. not null), as we have chosen $p$ to be the minimal such index s.t. $(T-lambda I)^p(x)=mathbf0$, i.e. $(T-lambda I)^p-1(x)neq mathbf0$. Note, that if $(T-lambda I)^p(x)=mathbf0$, then $(T-lambda I)^p+k(x)=mathbf0$ for any $k$ in the similar way as above.






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              The quoted part says something about the eigenvectors of $T$ when $p$ is the smallest positive integer such that $(T-lambda I)^p-1(x)$ equals zero. It is not merely saying that statement for the smallest positive integer, which is $p = 1$. However, it is true that if $p=1$, then $(T-lambda I)^p-1(x) neq 0$ for the reason you have stated (for $p = 1$, $(T-lambda I)^p-1(x) = x neq 0$ by assumption).



              In general, suppose $p$ is the least positive integer such that $(T-lambda I)^p(x) = 0$. Consider the vector $y = (T-lambda I)^p-1(x)$. We know that $y neq 0$, since if $y$ were equal to zero, then $p-1$ would be an integer smaller than $p$ with the property that $(T-lambda I)^p-1(x) = 0$, which contradicts that $p$ is the least such positive integer.
              Now, we write $$(T-lambda I)^p(x) = [(T-lambda I) circ (T-lambda I)^p-1](x) = (T-lambda I)[(T-lambda I)^p-1(x)] = (T-lambda)(y).$$
              Since $(T-lambda I)^p(x)=0$ by assumption, we have that
              $$
              0 = (T-lambda I)(y) implies 0 = Ty - lambda y implies Ty = lambda y.
              $$
              This shows that $y=(T-lambda I)^p-1(x)$ is an eigenvector of $T$ with eigenvalue $lambda$ when $p$ is the smallest positive integer such that $(T-lambda I)^p(x) = 0$.






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                3 Answers
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                We're saying that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$ then for some $n$ we have $(T-lambda I)^n(x) =0$.



                Note that this $n$ is not unique, because $(T-lambda I)^n+1(x) =0$ too. But we can choose $p$ to be the smallest such $n$ (because any non-empty set of natural numbers has a minimum element).



                Then if we set $v = (T-lambda I)^p-1(x)$ then $vneq 0$ by minimality of $p$, and we have $(T-lambda I)(v) = 0$, so $v$ is an eigenvector of $T$ corresponding to $lambda$.






                share|cite|improve this answer

























                  up vote
                  2
                  down vote



                  accepted










                  We're saying that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$ then for some $n$ we have $(T-lambda I)^n(x) =0$.



                  Note that this $n$ is not unique, because $(T-lambda I)^n+1(x) =0$ too. But we can choose $p$ to be the smallest such $n$ (because any non-empty set of natural numbers has a minimum element).



                  Then if we set $v = (T-lambda I)^p-1(x)$ then $vneq 0$ by minimality of $p$, and we have $(T-lambda I)(v) = 0$, so $v$ is an eigenvector of $T$ corresponding to $lambda$.






                  share|cite|improve this answer























                    up vote
                    2
                    down vote



                    accepted







                    up vote
                    2
                    down vote



                    accepted






                    We're saying that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$ then for some $n$ we have $(T-lambda I)^n(x) =0$.



                    Note that this $n$ is not unique, because $(T-lambda I)^n+1(x) =0$ too. But we can choose $p$ to be the smallest such $n$ (because any non-empty set of natural numbers has a minimum element).



                    Then if we set $v = (T-lambda I)^p-1(x)$ then $vneq 0$ by minimality of $p$, and we have $(T-lambda I)(v) = 0$, so $v$ is an eigenvector of $T$ corresponding to $lambda$.






                    share|cite|improve this answer













                    We're saying that if $x$ is a generalized eigenvector of $T$ corresponding to $lambda$ then for some $n$ we have $(T-lambda I)^n(x) =0$.



                    Note that this $n$ is not unique, because $(T-lambda I)^n+1(x) =0$ too. But we can choose $p$ to be the smallest such $n$ (because any non-empty set of natural numbers has a minimum element).



                    Then if we set $v = (T-lambda I)^p-1(x)$ then $vneq 0$ by minimality of $p$, and we have $(T-lambda I)(v) = 0$, so $v$ is an eigenvector of $T$ corresponding to $lambda$.







                    share|cite|improve this answer













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                    share|cite|improve this answer











                    answered 2 days ago









                    Daniel Mroz

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                        You can see this property by expanding the definition of exponentials of linear maps. This is nothing else than repeated composition.



                        You have $(T-lambda I)^p(x)=(T-lambda I)((T-lambda I)^p-1(x))=mathbf 0$, i.e. per definition $(T-lambda I)^p-1(x)inmathrmker(T-lambda I)$, i.e. $(T-lambda I)^p-1(x)$ is an eigenvector.



                        Now, this eigenvector really is proper(i.e. not null), as we have chosen $p$ to be the minimal such index s.t. $(T-lambda I)^p(x)=mathbf0$, i.e. $(T-lambda I)^p-1(x)neq mathbf0$. Note, that if $(T-lambda I)^p(x)=mathbf0$, then $(T-lambda I)^p+k(x)=mathbf0$ for any $k$ in the similar way as above.






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          You can see this property by expanding the definition of exponentials of linear maps. This is nothing else than repeated composition.



                          You have $(T-lambda I)^p(x)=(T-lambda I)((T-lambda I)^p-1(x))=mathbf 0$, i.e. per definition $(T-lambda I)^p-1(x)inmathrmker(T-lambda I)$, i.e. $(T-lambda I)^p-1(x)$ is an eigenvector.



                          Now, this eigenvector really is proper(i.e. not null), as we have chosen $p$ to be the minimal such index s.t. $(T-lambda I)^p(x)=mathbf0$, i.e. $(T-lambda I)^p-1(x)neq mathbf0$. Note, that if $(T-lambda I)^p(x)=mathbf0$, then $(T-lambda I)^p+k(x)=mathbf0$ for any $k$ in the similar way as above.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            You can see this property by expanding the definition of exponentials of linear maps. This is nothing else than repeated composition.



                            You have $(T-lambda I)^p(x)=(T-lambda I)((T-lambda I)^p-1(x))=mathbf 0$, i.e. per definition $(T-lambda I)^p-1(x)inmathrmker(T-lambda I)$, i.e. $(T-lambda I)^p-1(x)$ is an eigenvector.



                            Now, this eigenvector really is proper(i.e. not null), as we have chosen $p$ to be the minimal such index s.t. $(T-lambda I)^p(x)=mathbf0$, i.e. $(T-lambda I)^p-1(x)neq mathbf0$. Note, that if $(T-lambda I)^p(x)=mathbf0$, then $(T-lambda I)^p+k(x)=mathbf0$ for any $k$ in the similar way as above.






                            share|cite|improve this answer















                            You can see this property by expanding the definition of exponentials of linear maps. This is nothing else than repeated composition.



                            You have $(T-lambda I)^p(x)=(T-lambda I)((T-lambda I)^p-1(x))=mathbf 0$, i.e. per definition $(T-lambda I)^p-1(x)inmathrmker(T-lambda I)$, i.e. $(T-lambda I)^p-1(x)$ is an eigenvector.



                            Now, this eigenvector really is proper(i.e. not null), as we have chosen $p$ to be the minimal such index s.t. $(T-lambda I)^p(x)=mathbf0$, i.e. $(T-lambda I)^p-1(x)neq mathbf0$. Note, that if $(T-lambda I)^p(x)=mathbf0$, then $(T-lambda I)^p+k(x)=mathbf0$ for any $k$ in the similar way as above.







                            share|cite|improve this answer















                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 2 days ago


























                            answered 2 days ago









                            zzuussee

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                                The quoted part says something about the eigenvectors of $T$ when $p$ is the smallest positive integer such that $(T-lambda I)^p-1(x)$ equals zero. It is not merely saying that statement for the smallest positive integer, which is $p = 1$. However, it is true that if $p=1$, then $(T-lambda I)^p-1(x) neq 0$ for the reason you have stated (for $p = 1$, $(T-lambda I)^p-1(x) = x neq 0$ by assumption).



                                In general, suppose $p$ is the least positive integer such that $(T-lambda I)^p(x) = 0$. Consider the vector $y = (T-lambda I)^p-1(x)$. We know that $y neq 0$, since if $y$ were equal to zero, then $p-1$ would be an integer smaller than $p$ with the property that $(T-lambda I)^p-1(x) = 0$, which contradicts that $p$ is the least such positive integer.
                                Now, we write $$(T-lambda I)^p(x) = [(T-lambda I) circ (T-lambda I)^p-1](x) = (T-lambda I)[(T-lambda I)^p-1(x)] = (T-lambda)(y).$$
                                Since $(T-lambda I)^p(x)=0$ by assumption, we have that
                                $$
                                0 = (T-lambda I)(y) implies 0 = Ty - lambda y implies Ty = lambda y.
                                $$
                                This shows that $y=(T-lambda I)^p-1(x)$ is an eigenvector of $T$ with eigenvalue $lambda$ when $p$ is the smallest positive integer such that $(T-lambda I)^p(x) = 0$.






                                share|cite|improve this answer



























                                  up vote
                                  0
                                  down vote













                                  The quoted part says something about the eigenvectors of $T$ when $p$ is the smallest positive integer such that $(T-lambda I)^p-1(x)$ equals zero. It is not merely saying that statement for the smallest positive integer, which is $p = 1$. However, it is true that if $p=1$, then $(T-lambda I)^p-1(x) neq 0$ for the reason you have stated (for $p = 1$, $(T-lambda I)^p-1(x) = x neq 0$ by assumption).



                                  In general, suppose $p$ is the least positive integer such that $(T-lambda I)^p(x) = 0$. Consider the vector $y = (T-lambda I)^p-1(x)$. We know that $y neq 0$, since if $y$ were equal to zero, then $p-1$ would be an integer smaller than $p$ with the property that $(T-lambda I)^p-1(x) = 0$, which contradicts that $p$ is the least such positive integer.
                                  Now, we write $$(T-lambda I)^p(x) = [(T-lambda I) circ (T-lambda I)^p-1](x) = (T-lambda I)[(T-lambda I)^p-1(x)] = (T-lambda)(y).$$
                                  Since $(T-lambda I)^p(x)=0$ by assumption, we have that
                                  $$
                                  0 = (T-lambda I)(y) implies 0 = Ty - lambda y implies Ty = lambda y.
                                  $$
                                  This shows that $y=(T-lambda I)^p-1(x)$ is an eigenvector of $T$ with eigenvalue $lambda$ when $p$ is the smallest positive integer such that $(T-lambda I)^p(x) = 0$.






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                                    up vote
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                                    down vote










                                    up vote
                                    0
                                    down vote









                                    The quoted part says something about the eigenvectors of $T$ when $p$ is the smallest positive integer such that $(T-lambda I)^p-1(x)$ equals zero. It is not merely saying that statement for the smallest positive integer, which is $p = 1$. However, it is true that if $p=1$, then $(T-lambda I)^p-1(x) neq 0$ for the reason you have stated (for $p = 1$, $(T-lambda I)^p-1(x) = x neq 0$ by assumption).



                                    In general, suppose $p$ is the least positive integer such that $(T-lambda I)^p(x) = 0$. Consider the vector $y = (T-lambda I)^p-1(x)$. We know that $y neq 0$, since if $y$ were equal to zero, then $p-1$ would be an integer smaller than $p$ with the property that $(T-lambda I)^p-1(x) = 0$, which contradicts that $p$ is the least such positive integer.
                                    Now, we write $$(T-lambda I)^p(x) = [(T-lambda I) circ (T-lambda I)^p-1](x) = (T-lambda I)[(T-lambda I)^p-1(x)] = (T-lambda)(y).$$
                                    Since $(T-lambda I)^p(x)=0$ by assumption, we have that
                                    $$
                                    0 = (T-lambda I)(y) implies 0 = Ty - lambda y implies Ty = lambda y.
                                    $$
                                    This shows that $y=(T-lambda I)^p-1(x)$ is an eigenvector of $T$ with eigenvalue $lambda$ when $p$ is the smallest positive integer such that $(T-lambda I)^p(x) = 0$.






                                    share|cite|improve this answer















                                    The quoted part says something about the eigenvectors of $T$ when $p$ is the smallest positive integer such that $(T-lambda I)^p-1(x)$ equals zero. It is not merely saying that statement for the smallest positive integer, which is $p = 1$. However, it is true that if $p=1$, then $(T-lambda I)^p-1(x) neq 0$ for the reason you have stated (for $p = 1$, $(T-lambda I)^p-1(x) = x neq 0$ by assumption).



                                    In general, suppose $p$ is the least positive integer such that $(T-lambda I)^p(x) = 0$. Consider the vector $y = (T-lambda I)^p-1(x)$. We know that $y neq 0$, since if $y$ were equal to zero, then $p-1$ would be an integer smaller than $p$ with the property that $(T-lambda I)^p-1(x) = 0$, which contradicts that $p$ is the least such positive integer.
                                    Now, we write $$(T-lambda I)^p(x) = [(T-lambda I) circ (T-lambda I)^p-1](x) = (T-lambda I)[(T-lambda I)^p-1(x)] = (T-lambda)(y).$$
                                    Since $(T-lambda I)^p(x)=0$ by assumption, we have that
                                    $$
                                    0 = (T-lambda I)(y) implies 0 = Ty - lambda y implies Ty = lambda y.
                                    $$
                                    This shows that $y=(T-lambda I)^p-1(x)$ is an eigenvector of $T$ with eigenvalue $lambda$ when $p$ is the smallest positive integer such that $(T-lambda I)^p(x) = 0$.







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                                    share|cite|improve this answer



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                                    answered 2 days ago









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