# There exists a direct sum decomposition V=W⊕Z into two subspaces, with T the projection from V onto W along Z is equivalent T∘T=T..

Let V be a vector space, and T∈L(V). Show that the following statements (i), (ii) are equivalent:

(i) There exists a direct sum decomposition V=W⊕Z into two subspace, with T the projection from V onto W along Z.

(ii) T∘T=T.

any one can help with it, i have no idea with this question

• One direction is easy. For the other: what is the kernel of $T$? How about the image? Nov 16, 2018 at 22:15

(i) $$\Longrightarrow$$ (ii):

$$V = W \oplus Z; \tag 1$$

by definition means that

$$V = W + Z, \; W \cap Z = \{0\}; \tag 2$$

we note that the decomposition of any $$v \in V$$ into

$$v = w + z, \; w \in W, \; z \in Z, \tag 3$$

is unique, for if

$$w_1 + z_1 = w_2 + z_2, \tag 4$$

then

$$W \ni w_1 - w_2 = z_2 - z_1 \in Z; \tag 5$$

thus,

$$w_1 - w_2 = 0 = z_2 - z_1 \Longrightarrow w_1 = w_2, \; z_1 = z_2 \tag 6$$

as claimed; therefore, since $$w$$ is unambiguously determined by $$v$$, we may define a function

$$T:V \to W, \; T(v) = T(w + z) = w; \tag 7$$

we investigate the linearity of $$T$$: if

$$v = av_1 + v_2, \tag 8$$

we may uniquely write

$$v_1 = w_1 + z_1, \; v_2 = w_2 + z_2, \tag 9$$

whence

$$v = a(w_1 + z_1) + w_2 + z_2 = (aw_1 + w_2) + (az_1 + z_2) \in W + Z, \tag{10}$$

uniquely; it follows that

$$Tv = T(av_1 + v_2) = aw_1 + w_2 = aTv_1 + Tv_2, \tag{11}$$

establishing the linearity of $$T$$.

We compute

$$T^2(w + z) = T(T(w + z)) = Tw = T(w + z), \tag{12}$$

whence

$$T^2 = T. \tag{13}$$

(ii) $$\Longrightarrow$$ (i):

$$T^2 = T \Longrightarrow T(T - I) = T^2 - T = 0; \tag{14}$$

set

$$W = T(V); \tag{15}$$

then, via (14):

$$w \in W \Longrightarrow \exists v \in V, \; w = Tv; \; Tw = T^2v = Tv = w; \tag{16}$$

we see that $$T$$ fixes $$w$$ pointwise; it acts as the identity on the subspace $$W$$. We may also set

$$Z = (I - T)V; \tag{17}$$

then, again by (14),

$$z \in Z \Longrightarrow \exists v \in V, z = (I - T)v; \; Tz = T(I -T)v = 0 \Longrightarrow z \in \ker T; \tag{18}$$

likewise,

$$z \in \ker T \Longrightarrow Tz = 0 \Longrightarrow z = Iz - Tz = (I - T)z \Longrightarrow z \in (I - T)V = Z; \tag{19}$$

thus,

$$Z = \ker T; \tag{20}$$

now if

$$y \in Z \cap W, \tag{21}$$

we have

$$y = Tv, \; v \in V; \tag{22}$$

$$Ty = 0; \tag{23}$$

therefore, again invoking (14),

$$y = Tv = T^2v = T(Tv) = Ty = 0; \tag{24}$$

we have then shown that

$$Z \cap W = \{0\}; \tag{25}$$

finally, for $$v \in V$$,

$$v = Iv - Tv + Tv = (I - T)v + Tv \in Z + W; \tag{26}$$

(25) and (26) show that

$$V = W \oplus Z; \tag{27}$$

(16) and (20) show that $$T$$ is a projection onto $$W$$ "along $$Z$$". Thus we have (ii) $$\Longrightarrow$$ (i).

The implication (i) $$\Rightarrow$$ (ii) follows from the definition.

The implication (ii) $$\Rightarrow$$ (i) is not hard to verify once you know that $$W=\operatorname{im}T$$ and $$Z=\operatorname{ker}T$$.

• what is im T?.. Nov 16, 2018 at 23:31
• The image of $T$. You could also write it as $T(V)$. Nov 16, 2018 at 23:31