# Proving that $T + S: V \rightarrow V$ is the identity transformation when $N(T) \subseteq R(S)$ and $R(T) \subseteq N(S)$ and $T^2v = Tv, \ S^2v = Sv$

Notations:

$$N(T)$$: Null space of $$T$$, $$R(T)$$: Range of $$T$$

$$n(T)$$: Nullity of $$T$$, $$r(T)$$: Rank of $$T$$

$$S$$ and $$T$$ are both transformations from $$V$$ to $$V$$ such that $$T^2(v) = T(Tv) = Tv$$ and $$S^2(v) = S(Sv) = Sv$$ and:

$$N(T) \subseteq R(S) \\ R(T) \subseteq N(S)$$

I have to prove that $$T + S$$ is the identity transformation.

Firstly, I proved that $$N(T) = R(S)$$ and $$R(T) = N(S)$$ with the dimension theorem. (Assuming that $$V$$ is finite-dimensional.):

$$dimV = n(T) + r(T) = n(S) + r(S), \\ \Rightarrow r(S) - n(T) = r(T) - n(S)$$

The left hand side is bigger than or equal to zero, the right side is less than or equal to 0. So they are both equal to 0.

That means:

$$TS(v) = ST(v) = 0$$

I took the equation $$(T + S)(v) = w$$ and applied $$T$$ and $$S$$ which gave me these 2 equations:

$$Tw = T^2v + TSv = Tv + 0, \\ Sw = STv + S^2v = 0 + Sv \\ \Rightarrow Tv = Tw, \ Sw = Sv$$

But this doesn't help because that doesn't mean that $$w$$ and $$v$$ are equal. I'm stuck here and don't know how to prove that they are equal. I'm not sure if my method is useful or not but this is what I tried. (It doesn't look standard anyway.) I would also like to know how this could be proved for when $$V$$ is infinite dimensional.

This is true for $$V$$ of arbitrary dimension.

Let's look at $$T$$ on its own first. Because $$T$$ is idempotent, i.e., $$T^2v = Tv$$ for all $$v$$, we have $$N(T) \cap R(T) = \{0\}$$, since if $$v \in R(T)$$, so that $$v = Tw$$ for some $$w$$, we have $$Tv = T^2w = Tw = v$$, so $$v \not\in N(T)$$ unless $$v = 0$$. We also have that any $$v$$ can be written as $$v = a + b$$, where $$a = v - Tv \in N(T)$$ and $$b = Tv \in R(T)$$ and that this expression is unique: if $$a + b = a' + b'$$ with $$a, a' \in N(T)$$ and $$b, b' \in R(T)$$, then $$a - a' = b - b' \in N(T) \cap R(T) = \{0\}$$, so that $$a = a'$$ and $$b = b'$$. The jargon for this situation is to say that $$V$$ is the (internal) direct sum of $$N(T)$$ and $$R(T)$$ and one writes $$V = N(T) \oplus R(T)$$. (Apologies if you know these concepts already and don't need all this detail.)

Similarly we must have $$V = N(S) \oplus R(S)$$.

Now assume that $$N(T) \subseteq R(S)$$ and $$R(T) \subseteq N(S)$$. Then we must have $$N(T) = R(S)$$: for, suppose $$v \in R(S)$$, we can write $$v = a + b$$ for unique $$a \in N(T) \subseteq R(S)$$ and $$b \in R(T) \subseteq N(S)$$, so as $$V = N(S) \oplus R(S)$$, by the uniqueness property for $$N(S) \oplus R(S)$$ we must have $$a = v$$ and $$b = 0$$, so $$v = a \in N(T)$$; this gives us that $$R(S) \subseteq N(T)$$ and hence $$N(T) = R(S)$$. Similarly, we must have $$N(S) = R(T)$$.

Now, for any $$v \in V$$, we have $$T(v - Tv - Sv)= Tv - T^2v - TSv = 0$$ because $$T^2v = Tv$$ and $$R(S) = N(T)$$; similarly $$S(v - Tv - Sv) = 0$$. So $$v - Tv - Sw \in N(T) \cap N(S) = N(T) \cap R(T) = \{0\}$$, so $$v = Tv + Sv$$. Hence $$T + S$$ is the identity.

• Actually, thank you so much for the details in the beginning. I didn't know anything about direct sums and my brain kept thinking about somehow showing that $V$ is "made of" those 2 subsets. Thank you for further clarification on other dimensions as well! I believe that there's a little error in the end. I think you meant to write $v - Tv - Sv$ instead of $v - Tv - Tw$
– Zara
Oct 28, 2020 at 7:31
• Thanks for pointing out the typo. I've fixed it. Oct 28, 2020 at 12:42