Why is $(T-lambda I)^pT(x) = T(T-lambda I)^p(x)$?

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Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be an eigenvalue of $T$. The generalized eigenspace of $T$ corresponding to $lambda_1$ denoted $K_lambda$, is the subset of $V$ defined by



$K_lambda$ = $x in V: (T-lambda I)^p(x) = 0$ for some positive integer $p$.



To show that $K_lambda$ is $T$-invariant, consider any $x in K_lambda$. Choose a positive integer $p$ such that $(T-lambda)^p = 0$. Then



$(T-lambda I)^pT(x) = T(T-lambda I)^p(x) = T(0) = 0$




I wanted to know how we go from $(T-lambda I)^pT(x) = T(T-lambda I)^p(x)$?




linear-algebra linear-transformations




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  • 1




    Induction on $p$? Or write $T$ as $(T-lambda I)+lambda I$?
    – Lord Shark the Unknown
    2 days ago










  • Take both expressions and expand out $(T-lambda I)^p$ using the binomial theorem. After simplifying, you will find there are equal.
    – Mike Earnest
    2 days ago














up vote
0
down vote

favorite













Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be an eigenvalue of $T$. The generalized eigenspace of $T$ corresponding to $lambda_1$ denoted $K_lambda$, is the subset of $V$ defined by



$K_lambda$ = $x in V: (T-lambda I)^p(x) = 0$ for some positive integer $p$.



To show that $K_lambda$ is $T$-invariant, consider any $x in K_lambda$. Choose a positive integer $p$ such that $(T-lambda)^p = 0$. Then



$(T-lambda I)^pT(x) = T(T-lambda I)^p(x) = T(0) = 0$




I wanted to know how we go from $(T-lambda I)^pT(x) = T(T-lambda I)^p(x)$?




linear-algebra linear-transformations




share|cite|improve this question

















  • 1




    Induction on $p$? Or write $T$ as $(T-lambda I)+lambda I$?
    – Lord Shark the Unknown
    2 days ago










  • Take both expressions and expand out $(T-lambda I)^p$ using the binomial theorem. After simplifying, you will find there are equal.
    – Mike Earnest
    2 days ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite












Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be an eigenvalue of $T$. The generalized eigenspace of $T$ corresponding to $lambda_1$ denoted $K_lambda$, is the subset of $V$ defined by



$K_lambda$ = $x in V: (T-lambda I)^p(x) = 0$ for some positive integer $p$.



To show that $K_lambda$ is $T$-invariant, consider any $x in K_lambda$. Choose a positive integer $p$ such that $(T-lambda)^p = 0$. Then



$(T-lambda I)^pT(x) = T(T-lambda I)^p(x) = T(0) = 0$




I wanted to know how we go from $(T-lambda I)^pT(x) = T(T-lambda I)^p(x)$?




linear-algebra linear-transformations




share|cite|improve this question














Definition. Let $T$ be a linear operator on a vector space $V$, and let $lambda$ be an eigenvalue of $T$. The generalized eigenspace of $T$ corresponding to $lambda_1$ denoted $K_lambda$, is the subset of $V$ defined by



$K_lambda$ = $x in V: (T-lambda I)^p(x) = 0$ for some positive integer $p$.



To show that $K_lambda$ is $T$-invariant, consider any $x in K_lambda$. Choose a positive integer $p$ such that $(T-lambda)^p = 0$. Then



$(T-lambda I)^pT(x) = T(T-lambda I)^p(x) = T(0) = 0$




I wanted to know how we go from $(T-lambda I)^pT(x) = T(T-lambda I)^p(x)$?





linear-algebra linear-transformations





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









José Carlos Santos

111k1695172




111k1695172









asked 2 days ago









K.M

461312




461312







  • 1




    Induction on $p$? Or write $T$ as $(T-lambda I)+lambda I$?
    – Lord Shark the Unknown
    2 days ago










  • Take both expressions and expand out $(T-lambda I)^p$ using the binomial theorem. After simplifying, you will find there are equal.
    – Mike Earnest
    2 days ago












  • 1




    Induction on $p$? Or write $T$ as $(T-lambda I)+lambda I$?
    – Lord Shark the Unknown
    2 days ago










  • Take both expressions and expand out $(T-lambda I)^p$ using the binomial theorem. After simplifying, you will find there are equal.
    – Mike Earnest
    2 days ago







1




1




Induction on $p$? Or write $T$ as $(T-lambda I)+lambda I$?
– Lord Shark the Unknown
2 days ago




Induction on $p$? Or write $T$ as $(T-lambda I)+lambda I$?
– Lord Shark the Unknown
2 days ago












Take both expressions and expand out $(T-lambda I)^p$ using the binomial theorem. After simplifying, you will find there are equal.
– Mike Earnest
2 days ago




Take both expressions and expand out $(T-lambda I)^p$ using the binomial theorem. After simplifying, you will find there are equal.
– Mike Earnest
2 days ago










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Since $T$ commutes with both $T$ and $lambdaoperatornameId$, $T$ commutes with $T-lambdaoperatornameId$ and therefore it comutes with any power of $T-lambdaoperatornameId$.






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    up vote
    5
    down vote



    accepted










    Since $T$ commutes with both $T$ and $lambdaoperatornameId$, $T$ commutes with $T-lambdaoperatornameId$ and therefore it comutes with any power of $T-lambdaoperatornameId$.






    share|cite|improve this answer

























      up vote
      5
      down vote



      accepted










      Since $T$ commutes with both $T$ and $lambdaoperatornameId$, $T$ commutes with $T-lambdaoperatornameId$ and therefore it comutes with any power of $T-lambdaoperatornameId$.






      share|cite|improve this answer























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        Since $T$ commutes with both $T$ and $lambdaoperatornameId$, $T$ commutes with $T-lambdaoperatornameId$ and therefore it comutes with any power of $T-lambdaoperatornameId$.






        share|cite|improve this answer













        Since $T$ commutes with both $T$ and $lambdaoperatornameId$, $T$ commutes with $T-lambdaoperatornameId$ and therefore it comutes with any power of $T-lambdaoperatornameId$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        José Carlos Santos

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