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Let $T$ be a linear operator on a vector space $V$. A subspace $W$ is called $T$-admissible if

  • $W$ is $T$-invariant
  • if $f(T)\beta$ lies in $W$, then there exists a vector $w$ in W, such that $f(T) \beta = f(T) w$ ($f$ is a polynomial)

This definition is taken from Hoffman-Kunze. My question is whether the second condition sufficient for implying invariance? I doubt it, but haven't been able to find a counter. Could someone help with that.

Note : For infinite dimensional spaces, this is not sufficient. Take the integral operator on the set of polynomials and take the subspace to be that of constant multiples.

What would happen in finite dimensional space?

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  • $\begingroup$ I don't think that your infinite dimensional counterexample works $\endgroup$ Aug 16, 2022 at 15:48
  • $\begingroup$ @BenGrossmann By constant multiples I meant space of constant polynomials. So, if $p(x)\neq0$ is a polynomial of degree $n$ and $f$ is a polynomial of degree $m$, then $f(T)p$ would be a polynomial of degree $m+n > 0$. Hence, the only vectors which satisfy our second condition is the zero polynomial. And zero belong to our space of constant polynomials. But the space is not invariant. $\endgroup$
    – Kr Dpk
    Aug 16, 2022 at 16:00
  • $\begingroup$ That makes sense now, thanks for elaborating. I agree, that example is fine $\endgroup$ Aug 16, 2022 at 16:44

2 Answers 2

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Assume the field $F$ of scalars is infinite, such as $\mathbb C$ or $\mathbb R$.

Suppose $V$ is finite-dimensional over $F$ with dimension $n$.

Since $F$ is infinite, there is a $\lambda\in F$ such that $T-\lambda I_n$ is invertible.

Then $(T-\lambda I_n)V= V$. Hence $((T-\lambda I_n)V) \cap W= W$. The condition of admissibility implies that $((T-\lambda I_n)V)\cap W= ((T-\lambda I_n)W )\cap W$. Hence, $$(T-\lambda I_n)W\supseteq W.$$ Since both sides have the same finite dimension, they must be equal. $$(T-\lambda I_n)W= W.$$ which implies $TW\subseteq W$.

Please check Kr Dpk's answer for a simple counterexample when $F$ is finite.

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This is just a counterexample for finite fields, based on @ApassJack answer. Consider a two dimensional vector space $V$ over field $\mathbb{Z}_2$, having basis, $x$ and $y$. Consider the linear transformation $T$ which maps $$ T(x) = 0.x + 0.y $$ $$ T(y) = y $$ Consider the subspace $W$ spanned by $x+y$. Clearly this is not invariant under $T$. But since $T^2= T$, we need to check only the polynomials $I,T,T+I$. Checking these, we get that it does indeed satisfy our second condition, as $f(T) V \cap W = {0}$, where $f$ is any non-identity polynomial.

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