Prove that W is $T^{-k}$ invariant for every $k$

Question

Show that if $T\colon V\to V$ is an invertible linear transformation, and $W$ is a $T$-invariant subspace, then $W$ is also a $T^{-k}$ invariant subspace for every natural $k$.

My thought is to do this with induction but I am struggling with the process. Help would be appreciated.

Answer 1

$\def\inv{^{-1}}$ It's true, as Kyle pointed out, that finite dimensionality is essential, but only $W$ needs to be finite dimensional. Assume $W$ finite dimensional. Then $T_{|W} : W \to W$ is linear and injective, hence surjective. Thus for $w \in W$, there exists $w'\in W$ such that $T(w') = w$. Applying $T\inv$ to both sides of this equation gives $w' = T\inv(w)$. This shows $W$ is invariant under $T\inv$.

Exercise: Construct an example with $V$ infinite dimensional.

Answer 2

If $T,T':V\to V$ are linear transformations with $W$ an invariant subpace, then $T(T'(W))\subset T(W)\subset W$, so $T\circ T'$ has $W$ as an invariant subspace. If you are not comfortable working with image sets, you can take $w\in W$, and reason that $T'(w)\in W$, so $T(T'(w))\in W$.

If $T:V\to V$ is an invertible linear transformation with $W$ an invariant subspace of a finite dimensional vector space $V$. Decompose $V$ as $W\oplus W'$, by which I mean choose a basis of $W$ and extend the basis to a basis of $V$, the extra vectors are the basis of $W'$. With respect to this basis, the block matrix of $T$ looks like $$\begin{pmatrix}A&B\\0&C\end{pmatrix}$$ where $A,B,C$ are matrices, with $A,B$ square and invertible. That $0$ matrix represents the fact that $W$ is an invariant subspace. The inverse is $$\begin{pmatrix}A^{-1}&-A^{-1}BC^{-1}\\0&C^{-1}\end{pmatrix}$$ which means $W$ is still an invariant subspace for $T^{-1}$.

I'll leave the induction for $T^{-k}=(T\circ\dots\circ T)^{-1}$ to you.

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Finite-dimensionality is essential. Consider $V=\mathbb{R}^{\mathbb{Z}}$, the vector space of infinite sequences of real numbers going in both directions. Let $T:V\to V$ be the right-shift operator: $T(v)_i=v_{i-1}$. The subspace $W$ of all sequences $v$ with $v_i=0$ for all $i\leq 0$ is invariant under $T$. But, if $w$ is the vector with $w_1=1$ and $w_i=0$ for all $i\neq 1$, $w\in W$ but $T^{-1}(w)\not\in W$.