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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.

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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.

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    $\begingroup$ 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 $v - Tv - Tw$ $\endgroup$
    – Zara
    Oct 28, 2020 at 7:31
  • $\begingroup$ Thanks for pointing out the typo. I've fixed it. $\endgroup$
    – Rob Arthan
    Oct 28, 2020 at 12:42

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