Idempotent linear transformation from $V$ to $V$ is the direct sum of $\operatorname{range}(T)$ and $\operatorname{null}(T)$

Question

The problem statement is:

If $T\in\mathcal{L}(V)$ and $T^2=T$ (idempotent), prove that $V=\operatorname{range}(T)\oplus \operatorname{null}(T)$.

I'm not exactly sure where to start with this problem.

Answer 1

Hint: Notice that $v=Tv+(v-Tv)$...

Answer 2

We need to show that

$V = \text{Range}(T) + \text{Null}(T) \tag 1$

with

$\text{Range}(T) \cap \text{Null}(T) = \{0\}; \tag 2$

Now with our colleague Parcly Taxis we write, for $v \in V$,

$v = Tv + (I - T)v; \tag 3$

we may then use the hypothesis

$T^2 = T \iff T - T^2 = 0 \iff T(I - T) = (I - T)T = 0, \tag 4$

to see that for

$w = (I - T)v \tag 5$

we have

$Tw = T(I - T)v = 0, \tag 6$

that is,

$w \in \text{Null}(T); \tag 7$

thus we see that (1) binds; we have also established that

$\text{Range}(I - T) \subset \text{Null}(T); \tag 8$

we observe that

$(I - T)^2 = I - 2T + T^2 = I - 2T + T = I - T; \tag 9$

therefore with $w$ as in (7) we find

$w = w - Tw = (I - T)w = (I - T)^2w = (I - T)(I - T)w, \tag{10}$

and thus we have

$w \in \text{Range}(I - T), \tag{11}$

so that

$\text{Range}(I - T) = \text{Null}(T); \tag{12}$

thus to validate (2) we need show

$\text{Range}(T) \cap \text{Range}(I - T) = \{0\}; \tag {12}$

if

$w \in \text{Range}(T) \cap \text{Range}(I - T), \tag{13}$

then there are $x, y \in V$ such that

$Tx = w =(I - T)y, \tag{14}$

whence

$w = Tx = T^2x = T(I - T)y = 0; \tag{15}$

thus (2) is affirmed and with it

$V = \text{Range}(T) \oplus \text{Null}(T). \tag {15}$

Remark: It is worth noting that there is no restriction on $\dim V$ in the statement of this result; $\dim V = \infty$ is within the purview of this proposition. End of Remark.

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