Reversing Linear Transformations

Throughout this post, let V,W be vector spaces over some field \mathbb F and T : V \to W be a linear transformation.

Problem 1. For any subspace K \subseteq V, prove that T(K) is a subspace of W. In particular, the range T(V) of T is a subspace of W.

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Solution. For any x,y \in K and for any scalar c \in \mathbb F,

T(x) + T(y) = T(x+y) \in T(K),\quad cT(x) = T(cx) \in T(K).

Problem 2. For any subspace L \subseteq W, prove that T^{-1}(L) is a subspace of V. In particular, the kernel \mathrm{ker}(T):= T^{-1}(\{0\}) of T is a subspace of V.

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Solution. For any x,y \in T^{-1}(L), T(x),T(y) \in K. Since K is a subspace of W, for any scalar c \in \mathbb F,

T(x+y) = T(x) + T(y) \in L,\quad T(cx) = cT(x) \in L.

Problem 3. For any x, y \in V, prove that T(x) = T(y) if and only if x-y \in \mathrm{ker}(T).

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Solution. We have T(x) = T(y) \iff T(x) - T(y) = 0 \iff T(x-y) = 0.

Problem 4. For subspaces K_1, K_2 \subseteq W, prove that

K_1 +K_2 := \{x + y : x \in K_1, y \in K_2\}

is a subspace of W. Also, for any c \in \mathbb F, prove that cK_1 := \{cx : x \in K_1\} is a subspace of W.

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Solution. For any x_1+y_1, x_2+y_2 \in K_1 + K_2,

(x_1+y_1) + (x_2+y_2) = (x_1 + x_2) + (y_1 + y_2) \in K_1 + K_2.

Similarly, for any scalar c \in \mathbb F,

c(x_1+y_1) = c x_1 + c x_2 \in K_1 + K_2.

The proof for cK_1 is similar.

Problem 5. For subspaces K_1, K_2 \subseteq W, prove that

T^{-1}(K_1 + K_2) \supseteq T^{-1}(K_1) + T^{-1}(K_2) + \mathrm{ker}(T),

where equality holds if T is surjective.

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Solution. For the first claim, to prove (\supseteq), we note that for any x_i \in T^{-1}(K_i), x_3 \in \mathrm{ker}(T),

T(x_1+x_2+x_3) = T(x_1) + T(x_2) \in K_1 + K_2

so that x_1+x_2+x_3 \in T^{-1}(K_1 + K_2).

To prove (\subseteq) when T is surjective, fix x \in T^{-1}(K_1 + K_2). Then T(x) \in K_1 + K_2. Hence, there exists y_i \in K_i such that T(x) = y_1 + y_2.

Since T is surjective, find x_i \in V such that T(x_i) = y_i \in K_i \Rightarrow x_i \in T^{-1}(K_i). Then T(x) = T(x_1) + T(x_2) = T(x_1 + x_2). By Problem 3,

x_3 := x - (x_1 + x_2) \in \mathrm{ker}(T).

Hence, x = x_1 + x_2 + x_3 \in T^{-1}(K_1) + T^{-1}(K_2) + \mathrm{ker}(T).

Problem 6. For any scalar c \in \mathbb F and subspace K \subseteq W, prove that T^{-1}(cK_1) = c T^{-1}(K_1).

(Click for Solution)

Solution. We make the quick observation that

\displaystyle  x \in K \quad \iff \quad cx \in cK,

for any vector subspace K. This means for any x \in K_1,

T(x) \in cK_1 \quad \iff \quad T(c^{-1} x) \in K_1,

where c^{-1} := 1/c for brevity. The result follows by bookkeeping.

—Joel Kindiak, 26 Jan 25, 1826H

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