# Invariant Subspace

• Apr 3rd 2011, 01:32 PM
letitbemww
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.
• Apr 3rd 2011, 04:46 PM
Drexel28
Quote:

Originally Posted by letitbemww
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.
• Apr 3rd 2011, 09:53 PM
letitbemww
Quote:

Originally Posted by Drexel28
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?
• Apr 3rd 2011, 09:57 PM
Drexel28
Quote:

Originally Posted by letitbemww
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.