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In Titu Andreescu's Essential Linear Algebra: a problem solving approach I faced the following questions:

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How do I prove the underlined relation? I tried direct element chasing but then the relation seems trivial. If $\mathrm{Im}(T_1)+\mathrm{Im}(T_2)$ consists of all the elements of the form $Y_1+Y_2$ with $Y_1 \in \mathrm{Im}(T_1), Y_2 \in \mathrm{Im}(T_2)$ and $\mathrm{Im}(T_1+T_2)$ consists of those $Y$ for which $\exists X\in V: Y=(T_1+T_2)(X)=T_1(X)+T_2(X)$ but here $T_1(X)\in\mathrm{Im}(T_1)$ and $T_2(X)\in\mathrm{Im}(T_2)$ so if we follow this line of thought they should be equivalent. I cannot find my mistake.

And for number $2$ how do we get there? I tried finding information about the subadditivity of $\mathrm{dim}$ but didn't find anything.

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Here in the end it says that we only used the injectivity of $S_1$ and the surjectivity of $S_2$. However I feel like we actually used the surjectivity of $S_1$ to prove that $S_1(U)=V$ (number $1$ in the picture) and the biijectivity of $S_2$ to conclude that the two spaces have the same dimension (number $2$ in the picture). Is there a flaw in my reasoning or is this a typo?

Thanks in advance!

Edit 1: It seems that the second question in the first picture is actually Grassman's formula which is described in an earlier chapter of the book, so we are done with that.

Summand
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2 Answers2

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The image of $T_1+T_2$ consists of all vectors of the form $(T_1+T_2)v = T_1v + T_2v,$ which are of course contained in the set $\text{im }T_1 + \text{im }T_2 = \{T_1v + T_2w \mid v, w \in V\}.$

However, this inclusion may be proper. This is because the image of $T_1+T_2$ might not include a vector like $T_1v + T_2w.$ For instance, if $T_1(v) = v$ and $T_2(v) = -v,$ then $T_1+T_2$ is the zero map, but $T_1, T_2$ each have image $V,$ so the sum of their images is $V.$

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$T_1 v + T_2 w$ is an element of $\text{Im}(T_1) + \text{Im}(T_2)$, but might not be in $\text{Im}(T_1 + T_2)$ when $v \ne w$.

For the dimension inequality, note that $\dim(U+V) = \dim(U) + \dim(V) - \dim(U \cap V)$ in general.

For your second question, you are correct that bijectivity of $S_1$ and $S_2$ are necessary.

Example to show that "$S_1$ is injective and $S_2$ is surjective" is not enough: Suppose $S_1 : \mathbb{R} \to \mathbb{R}^2$ is injective and $S_2 : \mathbb{R}^2 \to \mathbb{R}$ is surjective, and $T: \mathbb{R}^2 \to \mathbb{R}^2$ is the identity. Then $\text{rank}(T)=2$ while $S_2 T S_1 : \mathbb{R} \to \mathbb{R}$ has rank at most $1$.

Response to comment: I now realize that the author probably meant "injectivity of $S_2$ and surjectivity of $S_1$." You have already noted that for $S_1$ only only surjectivity was used. For $S_2$, note that injectivity is sufficient to justify "$(TS_1)(U)$ and $S_2 ((T S_1)(U))$ are isomorphic." Here, $S_2$ is mapping a subspace of $W$ to a subspace of $Z$, and it is not necessary for $S_2$ to map $W$ onto all of $Z$.

angryavian
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  • In your example the biijectivity of both maps is indeed necessary and I see how it is used, but in the proof of the book I still cannot see the where the necessity of the injectivity of $S_1$ is used (I see the surjectivity is used for the equality number $1$). – Summand Jul 06 '21 at 23:49
  • @Summand See my update – angryavian Jul 07 '21 at 00:00
  • You said: For S2, note that injectivity is sufficient to justify "$(TS_1)(U)$ and $S_2((TS_1)(U))$ are isomorphic." However doesn't injectivity only guarantee that $\mathrm{dim}((TS_1)(U)) \leq \mathrm{dim}(S_2((TS_1)(U)))$? I feel like surjectivity is also necessary but I wanna confirm. – Summand Jul 07 '21 at 00:09
  • @Summand Yes. The general statement is "if $S : V \to W$ is an injective linear map, then $S$ is an isomorphism between $V$ and its image $S(V)$." (If $S$ is surjective then $S(V)=W$. If $S$ is not surjective, then $S(V)$ is a strict subspace of $W$.) You can probably prove this by working with a basis or something. – angryavian Jul 07 '21 at 00:58