Let $V$ be a vector space over $\Bbb R$ of dimension $n$, and $T \colon V \to V$ be a linear trasformation. Choose the correct answer.
(a) There exist subspaces $V := V_0 \subset V_1 \subset V_2 \dots \subset V_n = V$ such that $T(V_i) \subset V_i$ and $\dim_{\mathbb R}(V_i)=i$.
(b) If $T$ is not surjective, then there exists a subspace $U \neq 0$ such that and $T(U)=0$.
(c) If there is a proper non-zero subspace $U$ such that $T(U) \subset U$, then there is a proper non-zero subspace $W$ such that $T(W) \subset W$ and $V = U \oplus W$.
(d) If $T(U) \cap U \neq 0$ for any subspace $U \neq 0$, then $ST=TS$ for all linear transformations $S$ from $V$ to $V$.
My attempt:
I started with (b) first, since it gave me the idea that $\text{Null}(T) \neq 0$ (as it is not onto, so is not one-to-one) and the kernel will surely be non trivial and nullity will be positive (so blindly I choose it correct).
For option (a) we have that there is the existence of $T$-invariant subspaces, viz. $V_i$ (but I think I couldn’t decode it properly).
For (c) take $U=W=(x,0)$ as a counter, since no comments are made for both of them being distinct.
For (d) define $T,S \colon V \to V$ where $V=\mathbb R^2$. Now for $U=(x,y)$, consider $S(x,y)=(-y,x)$ and $T(x,y)=(x,0)$; then $ST(x,y)=(0,x)$ and $TS(x,y)=(0,0)$.
More counters invited :~)
Any help with option a or any correction in above try that I made? Thanks.
(a) is wrong in general as it implies that it exists a basis in which the matrix of $T$ is upper triangular and as a consequence that the characteristic polynomial of $T$ is a product of linear factors. Which may not be the case. By the way (a) is equivalent to the fact that $T$ is similar to an upper triangular matrix.
(c) Your counterexample is quite strange as you don’t define $T$. A counterexample is provided by
$$T= \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}$$ as you can prove that $\mathbb R e_1$ is the only proper stable subspace.