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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

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

2 Answers 2

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(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).

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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 $Z=\operatorname{ker}T$.

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

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