Understanding of basic properties of linear maps I'm studying for my Linear Algebra exam, and this is one of the questions in a previous exam. Sadly I'm not quite sure I get it.
The question goes:
Let $V$ be a finite-dimensional $R$−vector space
Let $U$ and $W$ be subspaces of $V$, and $T,S \in\mbox{End}(V)$. Decide if the statements are true or false for any choice of $U, W, T$, and $S$.


*

*$\dim(U + W) = \max\{\dim(U),\dim(W)\}$

*$\dim(U + W) ≥ \min\{\dim(U),\dim(W)\}$

*$\dim(U\oplus W) =\dim(U) + \dim(W)$

*$\dim(U \cap W) = \min\{\dim(U), \dim(W)\}$

*$\{0\}\subseteq U\cap W$

*$U \cup W$ is a subspace of $V$

*Let $\{v_1, . . . , v_n\}$ be a set of linearly independent vectors in $V$.
Then $\dim(V) ≥ n$.

*$\dim(\ker(T)) + \dim(\mbox{im}(T)) = \dim(V )$

*There exists $F\in\mbox{Hom}(R^2 , R^3)$ such that $F$ is injective.

*$\mbox{im}(ST) = \mbox{im}(S)$
Course vs. Book
$\mbox{Hom}(V, W) | L(V, W)$
$\mbox{End}(V ) | L(V )$
$\ker |\mbox{null}$
$\mbox{im}| \mbox{range}$
My thoughts:


*

*So I guess this is false, cause $dim(U+W)$ are suppose to be equal OR bigger
than $\max\{\dim(U), dim(W)\}$, but theoretically I don't understand why?

*Must be True, cause the absolute minimum of the two dimensions, will always be smaller than the dimension of their addition?

*This I've been told is True, because $dim(U \oplus W) = dim(U) + dim(W)-dim(U\cap W)$ but why is it that the intersection is $0$? 

*This again, I've been told that the intersection is $0$ and the minimum is $1$, but I'm not sure why?

*Well if the intersection is zero, then, of course, {$0$} must be in the intersection.

*Must be false, cause units are not subspaces.

*True. Of course, there could not be a vector bigger than the dimension, but the dimension does contain smaller vectors. (This is not good math language I know, just my thought to understand it.)

*This we know is true from the Fundamental Theorem of Linear Maps ("Linear Algebra Done Right" by Sheldon Axler)

*I do believe this is true, but can't explain it?

*I don't know, but I don't get how the range of two subspaces could be the same as one of them? Maybe if T is also a subspace of S? Not sure.
 A: You have done well. But to get full marks in an exam you need to give a counterexample to prove a statement false or a short proof (or sometimes example) to prove a statement true.
(1) Correct. It is false. Counterexample: $U$ = all multiples of $\mathbf{u}$ and $V$ = all multiples of $\mathbf{v}$ where $\mathbf{u,v}$ are linearly independent. Then dim($U$)=dim($V$)=1 but dim($U+V$)=2.
(2) Correct. True because $U\subseteq U+V$.
(3) Correct. True. By definition $U\oplus V$ has $U\cap V=0$.
(4) Correct. False. Take the same $U,V$ as in (1) above. The dim($U$)=dim($V$)=1 but dim($U\cap V$)=0.
(5) Correct. True. The zero vector belongs to any subspace, so it must belong to $U\cap V$.
(6) Correct. False. Take $U,V$ as in (1) above. Then $\mathbf{u}+\mathbf{v}\notin U\cup V$, but $\mathbf{u,v}\in U\cup V$. So $U\cup V$ is not closed under vector addition.
(7) Correct. True. The dimension is the maximum number of linearly independent vectors.
(8) Correct. True. The rank-nullity theorem.
(9) Correct. True. Example: define $F(x,y)=(x,y,0)$. [Check this is a homomorphism and injective].
(10). False. I do not know whether you take $ST$ to mean first $S$, then $T$ or vice versa (both conventions are used). But either way take $S$ to be the identity map and $T$ the null map. Then the image of $S$ is $V$, but the image of $ST$ is 0.
------- Added later -------
re (9) and your comment below. Yes. Injective means one-one, or as you say, $Tu=Tv$ implies $u=v$. That is clearly true for the example given above.
re (9) and your second comment below. To check that $T$ is a homomorphism we just need it to be a linear map, ie $T(\lambda u+\mu v)=\lambda Tu+\mu Tv$ for all vectors $u,v$ and scalars $\lambda,\mu$. In the case of the example above that is obviously true.
