Suppose that $V$ is a $4-dimensional$ vector space over $R$. Let $T :V → V$ be an invertible linear map and $W$ be a proper subspace of $V$ , such that $V = W ⊕ TW$. If $T^2W ⊆ W$, then show that $T^2W = W$.
Since $V = W ⊕ TW$, so for $v ∈ V$ uniquely as $v = w + Tw$, where $w ∈ W$ and $Tw ∈ TW$.
Let $w ∈ W$. Since $V = W ⊕ TW$, we can write $w = w + 0T$, where $0T$ is the zero vector in $TW$. Since $T$ is invertible,
we have $T^{-1}(w + 0T) = T^{-1}(w) + T^{-1}(0T) = T^{-1}(w)$,
which means that $w = T(T{^-1}(w)) ∈ T(W)$.
Thus, we have shown that $W ⊆ T(W)$. Since T is invertible,
we have $V = T(V) = T(W ⊕ TW) =TW⊕T^2W$,
so $V = W ⊕ TW ⊕ T^2W$.
Now, let $v ∈ T^2W$. Since $T^2W ⊆ W$, we have $v ∈ W$, and since $V = W ⊕ TW ⊕ T^2W$, we can write $v$ uniquely as $v = w_1 + w_2$, where $w_1 ∈ TW$ and $w_2 ∈ T^2W$. Since $T^2W$ is a subspace of $V$ and $T$ is linear, we have $T(w_2) ∈ T(T^2W) = T^3W ⊆ T^2W$. Therefore, $T(w_2)$ can be written as a linear combination of vectors in $T^2W$, say $T(w_2) = a_1v_1 + a_2v_2 + ... + a_nv_n$, where $v_1, v_2, ..., v_n$ are vectors in $T^2W$ and $a_1, a_2, ..., a_n$ are scalars. Since $T$ is invertible, we have $T^{-1}(v_1), T^{-1}(v_2), ..., T^{-1}(v_n) ∈ T(T^2W) = T^3W ⊆ T^2W$.
Thus, we can write $T^{-1}(v_i) = b_{1i} v_1 + b_{2i} v_2 + ... + b_{mi} v_m$, where $v_i, v_2, ..., v_m$ are vectors in $T^2W$ and $b_{1i}, b_{2i}, ..., b_{mi}$ are scalars.
Now, we have:
$w_2 = T^{-1}(T(w_2)) = T^{-1}(a_1v_1 + a_2v_2 + ... + a_nv_n)$
= $b_{11}v_1 + b_{12}v_2 + ... + b_{1m}v_n + b_{21}v_1 + b_{22}v_2 + ... + b_{2m}v_n + ... + b_{n1}v_1 + b_{n2}v_2 + ... + b_{nm}v_n$
Since each $v_i ∈ T^2W$ and $T^2W$ is a subspace of $V$, we have $w_2 ∈ T^2W$. Therefore, $v = w_1 + w_2 ∈ T^2W$, which shows that $T^2W ⊆ W$.
Since we already showed that $W ⊆ T^2W$, we have $T^2W = W$, as desired.
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