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Math Help - Invariant Subspace

  1. #1
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    Invariant Subspace

    If T is an upper triangular matrix and Fk is the subspace of Rn defined as Fk = span{e1,...ek} where ei is the standard basis vector, show that Fk is an invariant subspace of T.

    Then, if we restrict T to be a strictly upper triangular matrix (so the diagonal is also 0), then show that T(Fk) is contained in F(k-1).

    I'm really just very confused about how to prove that something is an invariant subspace. Also, isn't the span of the standard basis vector just Rn?

    I have that if v is in Fk, then v=sum(i=1 to k)(ai)(ei) for some scalars ai. Applying T and using linearity gives Tv=sum(i=1 to k)(ai)(Tei). I just don't know where to go from there.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by letitbemww View Post
    If T is an upper triangular matrix and Fk is the subspace of Rn defined as Fk = span{e1,...ek} where ei is the standard basis vector, show that Fk is an invariant subspace of T.

    Then, if we restrict T to be a strictly upper triangular matrix (so the diagonal is also 0), then show that T(Fk) is contained in F(k-1).

    I'm really just very confused about how to prove that something is an invariant subspace. Also, isn't the span of the standard basis vector just Rn?

    I have that if v is in Fk, then v=sum(i=1 to k)(ai)(ei) for some scalars ai. Applying T and using linearity gives Tv=sum(i=1 to k)(ai)(Tei). I just don't know where to go from there.
    So by assumption we may write T as \displaystyle \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}. So, if e_\ell stands for the \ell^{\text{th}} canonical basis vector for \mathbb{R}^n then the question says to define F_k=\text{span}\{e_1,\cdots,e_k\}. But, by definition for each \displaystyle \sum_\ell \alpha_\ell e_\ell\in F_k we have that


    \displaystyle T\left(\sum_\ell \alpha_\ell e_\ell\right)=\sum_\ell \alpha_\ell T(e_\ell)=\sum_\ell \alpha_\ell \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}e_\ell


    But write out \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}e_\ell and I'm sure you'll see that it's in \text{span}\{e_1,\cdots,e_\ell\}\subseteq\text{spa  n}\{e_1,\cdots,e_k\} so that each element of summand is in \text{span}\{e_1,\cdots,e_k\} and thus so is there sum, etc.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    So by assumption we may write T as \displaystyle \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}. So, if e_\ell stands for the \ell^{\text{th}} canonical basis vector for \mathbb{R}^n then the question says to define F_k=\text{span}\{e_1,\cdots,e_k\}. But, by definition for each \displaystyle \sum_\ell \alpha_\ell e_\ell\in F_k we have that


    \displaystyle T\left(\sum_\ell \alpha_\ell e_\ell\right)=\sum_\ell \alpha_\ell T(e_\ell)=\sum_\ell \alpha_\ell \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}e_\ell


    But write out \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{pmatrix}e_\ell and I'm sure you'll see that it's in \text{span}\{e_1,\cdots,e_\ell\}\subseteq\text{spa  n}\{e_1,\cdots,e_k\} so that each element of summand is in \text{span}\{e_1,\cdots,e_k\} and thus so is there sum, etc.
    Oh thanks. That makes sense. So for the second part with the strictly upper triangular matrix, would I just substitute the matrix you gave for T with a matrix that has zero on the main diagonal, and that would make T basically shifted over one column? And that would make T(e_i) in span(e_1,...,e_l-1) which is just in span(e_1,...,e_k-1), so this is just in F_k-1?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by letitbemww View Post
    Oh thanks. That makes sense. So for the second part with the strictly upper triangular matrix, would I just substitute the matrix you gave for T with a matrix that has zero on the main diagonal, and that would make T basically shifted over one column? And that would make T(e_i) in span(e_1,...,e_l-1) which is just in span(e_1,...,e_k-1), so this is just in F_k-1?
    Basically--yes.
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