$T$-admissible subspace: an equivalent way?

Question

In the book of Linear Algebra by Hoffman-Kunze, the authors write

Let $T$ be a linear operator on a vector space $V$ and let $W$ be a subspace of $V$. We say that $W$ is $T$-admissible if

1. $W$ is invariant under $T$;

2. if $f(T)\beta$ is in $W$, then there exists a vector $\gamma$ in $W$ such that $f(T)\beta=f(T)\gamma$ (whete $f(T)$ is any polynomial in $T$).

Question: Can we state the above conditions in following form: $$T(V)\cap W=T(W)? \hskip1cm (*)$$ From (1) and (2), it is easy to verify this condition; I am not sure whether $(*)$ implies (1) and (2)?

Any hint or suggestion?

Answer 1

No, $(*)$ does not imply that $W$ is $T$-admissible. As a counterexample, take $V = \Bbb R^2$, $W = \{(x,0): x \in \Bbb R\}$, and $$ T = \pmatrix{0&1\\0&0}. $$ Note that condition 2 fails with $\beta = (0,1)$ and $f(x)=x.$