Linear transformation and direct sum

Question

When given $T:V \to V$ such that $T^2=I$, prove that $U,W \subseteq V$ subspaces exist such that:

$$V=U\oplus W$$

$$T(u)=u, \forall u \in U$$

$$T(w)=-w, \forall w \in W$$

My first thought was to define $W=kerT$ and $U=ImT$, then I can prove that $\dim(U+W)=\dim(U)+\dim(W) = \dim(V)$. that shows $U \oplus W=V$, and also $T(-w)=w=\bar{0}$.

But the second condition is not met. I was also trying to define $U=sp\{\bar{u_1},....,\bar{u_n}\}\subseteq V$ as a subspace but I can't prove that $T(u)=u$

The problem should be solved with knowledge about direct sums and linear transformation.

any help will be much appreciated

Answer 1

As I am just a student take carefully my solution and I hope it is correct.
I think that there is a more elegant way than a prove by contradiction to prove (2). (btw i hope too that i ve understood correctly your question).

1 - Prove that $U$ and $W$ are in direct sum.
$U=\left \{ u \in V: T(u)=u \right \}$ and $W=\left \{ w \in V: T(w)=-w \right \}$.
It cames that any $v' \in U \cap W $ has to verify: $T(v')=v'$ as $v' \in U$ and in the same time $T(v')=-v'$ as $v' \in W$.
Moreover by definition of application the same input always give the same output so we have: $v'=T(v')=-v' \Rightarrow v'=0_V$ is the unique solution. Hence: $U \cap W = 0_V \Rightarrow U\oplus W$

2 - Prove that $V=U+W$.
By absurd we suppose $\exists v' \in V\setminus \left \{ U + W \right \}$.
Now let's note $T(v')=v'' \neq \pm v' \Rightarrow T(T(v')) = T(v'') = v' \neq \pm v''$. In this case it cames that: $(v'+v'')/2 \in U$ and $(v'-v'')/2 \in W$. So by definition as the sum of one element in $U$ and one element in $W$ we get: $(v'+v'')/2+(v'-v'')/2=v' \in U+W$ contradiction.

(1) & (2) $\Rightarrow$ Q.E.D.

Answer 2

We need to prove that $U \cap W = \{0\}$ and that $ U + W = V$.

The first is easy to solve. Suppose that $v \in U \cap W$ then we have $T(v) = v$ and $T(v) = -v$ so $v$ must be equal to $0$.

For the second part let $z \in V$. Then

$$ \frac{z + T(z)}{2}\in U$$ $$ \frac{z - T(z)}{2}\in W.$$

since $T^2 = I$. It follows that

$$ \underbrace{\frac{z + T(z)}{2} + \frac{z - T(z)}{2}}_{ = z} \in U + W.$$

Hence $V \subset U + W$ which proves $V = U + W$.

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In that question we are working over some field F. I guess it is not causing problems, I just wonder what meaning $\frac{1}{2}$ have when we are working over some abstract field.

If $V$ is a vector space over a field $ F$ then we can multiply vectors $v$ by scalars $\lambda \in F$. When we write $$ v + v = 2v$$ then the number $2$ is not necessarily the number $2 \in \mathbb N$. Denote by $1_F$ the neutral element for multiplication in the field $F$. Then

$$ v + v = 1_Fv + 1_Fv =(1_F + 1_F )v $$

so when we write $2$ we actually mean $1_F + 1_F$. In particular $\frac{1}{2}$ denotes the multiplicative inverse of $2$ in the field $F$. But can we be sure that $2$ is actually invertible in $F$ ? Recalling the properties of a field you know that $\lambda \in F$ is invertible if and only if $\lambda \neq 0_F$. So as long as you are working in field $F$ such that $2 \neq 0_F$ then the above proof is valid. That is, as long as $1_F + 1_F \neq 0_F$ you wont have a problem.

It might be surprising but there actually exists fields $F$ such that $1 + 1 = 0$. For example the finite field $\mathbb F_2$.