3
$\begingroup$

Let the linear transformation T on the vector space $V$ over $\mathbb{Q}$ have minimal polynomial $(x^{7} - x^{3})$.

a) What is the largest invariant subspace W of V for which $T (W) = W$?

b) Find a non-constant polynomial $f(x) \in \mathbb{Q} [x] $ for which $f(T)w = w$, for all $w \in W$.

For part (a): If $W$ is any invariant subspace of T, then as an application of Division Algorithm the minimal polynomial of $T|_{W}$ divides $m_{T}$ and using this helps one to show the existence of a polynomial $p(x)$ over $\mathbb{Q}$ such that $p(x)|m_{T}(x) = (x^{7} - x^{3})$ and $W = \ker (p(T))$. Since $p(T)W = 0$, the requirement that $TW = W$ is met when $W = \ker (T - I)$. I guess what is meant by largest refers to dimension. Here, I hope I have found an invariant subspace that works, but is it the largest? I would appreciate a hint.

For part b): The polynomial $f(x) = x - 1$ satisfies the requirement.

$\endgroup$
1
$\begingroup$

Your question (a) is a bit strangely stated, but it actually asks for the largest invariant subspace for which its operator given by restricting $T$ to it is surjective (since $T(W)\subseteq W$ must hold by definition). Given that the minimal polynomial is $X^7-X^3=(X^4-1)X^3$ one can see that $T$ itself (acting on all of$~V$) is not surjective: the image $T^3(V)$ is contained in the kernel of $\def\I{\mathbf I}T^4-\I$, which is not the whole space (because $T^4-\I\neq0$), but if would have to be if $T$ were surjective.

By a small variation of that argument, any (invariant) subspace $W$ such that $T(W)=W$ must be contained in $\ker(T^4-\I)$, namely $\{0\}=(T^4-\I)\circ(T^3(W))=(T^4-\mathbf I)(W)$. But the subspace $W_0=\ker(T^4-\I)$ itself is $T$-stable, and $T|_{W_0}$ is invertible since its $4$-th power is the identity of $W_0$, so $W_0=\ker(T^4-\I)$ must be the answer to the first question.

The second question is quite independent. Your answer $X-1$ is wrong, since if it acted like $\I$, that is if one had $T-\I=\I$, this would mean $T=2\I$ which is incompatible with the given minimal polynomial. Of course the polynomial $1$ does act as $\I$, but a non constant polynomial is asked for. We must add to $1$ some nonzero polynomial that evaluated in $T$ does act as zero. Clearly the given minimal polynomial is such a polynomial; therefore $X^7-X^3+1$ answers the second question (or you could add any nonzero multiple of the minimal polynomial to $1$ for alternative answers).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.