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$V$ is finite-dimensional with $\dim V>1$ and $T\in L(V)$. Then $$\{p(T)\,|\,p\in p(\mathbb F)\} \ne L(V)$$

When I see this problem, firstly I do not believe in above as $L(V)$ is vector space then any linear combination of vector in itself must be in that.

But after that I thought it may be possible that there exist some more elements in $L(V)$ than that polynomial. I know that $\{p(T)\,|\,p\in \mathbb F(x)\}$ contain elements which are commutative. That means I have to find some non commutative elements in $L(V)$.

That is I can always find as $ST\neq TS$ where $$S(e_1)=e_2,S(e_2)=e_1,S(e_i)=e_i \text{ for } i=3,4,\ldots, n$$ and $$T(e_1)=e_1,T(e_2)=e_1,T(e_i)=e_i$$ Is above opinion about problem is right? Or where is my mistake in arguing?

Any help will be appreciated

mechanodroid
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Curious student
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2 Answers2

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The space $L(V)$ is of dimension $n^2$ where $n=\dim V$.

By Cayley–Hamilton theorem, $\dim W \le n$ where $W=\{p(T) \colon p\in \mathbb F[x] \}$.

As $n$ is supposed to be greater than one you have $n^2 >n$ and $W \subsetneq L(V)$.

mathcounterexamples.net
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Recall that an operator $T \in L(V)$ commutes with all operators from $L(V)$ if and only if $T$ is a scalar multiple of the identity, i.e. $T = \lambda I$.

If $T = \lambda I$, then $\{p(T) : p \in \mathbb{F}[x]\}$ also consists only of scalar multiples of the identity so clearly $\{p(T) : p \in \mathbb{F}[x]\} \ne L(V)$.

If $T$ is not a scalar multiple of the identity, then there exists $S \in L(V)$ such that $TS \ne ST$. Clearly $T$ commutes with all $p(T)$ for $p \in \mathbb{F}[x]$ so $S \notin \{p(T) : p \in \mathbb{F}[x]\}$.

mechanodroid
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