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Let $V$ be a vector space and $T:V\rightarrow V$ a linear transformation with the property that $T(W)\subseteq W$ for every subspace $W$ of $V$. Prove that $T$ is scalar multiplication, i.e. there is an element $\lambda$ in the field of scalars such that $T(v)=\lambda v$ $\forall v\in V$.

My attempt: I gather that for any element $w$ in a subspace $W$ with basis $\{w_1,\dots,w_n\}$, we have

$w = a_1w_1+\dots+a_nw_n$

for scalars $a_1,\dots,a_n$.

We also know that $T(w) = T(a_1w_1)+\dots+T(a_nw_n)$, and that since for each $i$, span$\{w_i\}$ is a subspace, $T(w_i)=\alpha_i(w_i)$ for some scalar $\alpha_i$.

I feel like this should be enough for the solution, but I can't get there.

Any help appreciated!

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    $\begingroup$ The statement doesn't seem to mention that the space is finite-dimensional, so you can't assume you have a finite basis. $\endgroup$ – Wojowu Mar 28 '17 at 18:32
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You have that $T(w_i)=\alpha_iw_i$. Now, note that for $w=w_1+\cdots+w_n$, we must have $T(w)=\beta w$ (since $\langle w\rangle$ is $T$-invariant). But, $$ \beta w_1+\cdots+\beta w_n=\beta w = T(w_1+\cdots+w_n)=T(w_1)+\cdots+T(w_n)=\alpha_1w_1+\cdots+\alpha_nw_n. $$ Hence $$ 0=(\alpha_1-\beta)w_1+\cdots+(\alpha_n-\beta)w_n. $$ As $\{w_i\}$ forms a basis we must have $0=\alpha_1-\beta=\cdots=\alpha_n-\beta$, or $\alpha_1=\cdots=\alpha_n=\beta$.

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Think of the subspaces $W_i= \langle w_i\rangle$. $T$ has to act by scalar multiplication on these spaces. Again consider the subspace $W_i+W_j$ and explore it.

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  • $\begingroup$ I think instead of $W_i+W_j$ you want to look at $\langle w_i+w_j\rangle$, is that what you meant by a chance? $\endgroup$ – Wojowu Mar 28 '17 at 18:27
  • $\begingroup$ But $\langle w_i+w_j \rangle $ is a subspasce of $W_i+W_j$ $\endgroup$ – Parish Mar 28 '17 at 18:28
  • $\begingroup$ No, they aren't. $W_i+W_j$ contains $w_i+0=w_i$, but $\langle w_i+w_j\rangle$ doesn't. $\endgroup$ – Wojowu Mar 28 '17 at 18:30
  • $\begingroup$ @Wojowu: Sorry, I had a brain fade, I have edited my previous comment. $\endgroup$ – Parish Mar 28 '17 at 18:32
  • $\begingroup$ Okay, it is a subspace. But that $T(W_i+W_j)\subset W_i+W_j$ doesn't imply $T(\langle w_i+w_j\rangle)\subseteq \langle w_i+w_j\rangle$. $\endgroup$ – Wojowu Mar 28 '17 at 18:36

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