Key to "Linear transformation and its"

cyrsvab

Answered question

2022-02-14

Linear transformation and its

Answer & Explanation

venenolundicoofw

Beginner2022-02-15Added 12 answers

First, the statement is false if $t = 0 in \frac{R^{p}}{M}$ because in this case, t is the class of an element t′ of M and $M = k e r B \Rightarrow B t^{'} = 0 \neq 1$ .
Now, if $t \neq 0$ , then $t^{'} ⧸ \in M$ and the line $L = ⟨ t^{'} ⟩$ satisfies $M \oplus L = R^{p}$ . Thus if $π$ is the projection on L along M (i.e., if $x = x_{M} + x_{L}$ , then $π (x) = x_{L})$ and if $ψ : L \to R$ is defined by $ψ (λ t^{'}) = λ, \forall λ \in R$ , we see that $B = ψ \circ π$ satisfies $k e r B = M and B t = 1$ .
Edit: 1. B is linear: indeed, if $x = x_{M} + λ t^{'} \in R^{p}, y = y_{M} + μ t^{'} \in R^{p} and a \in R then a x + y = (a x_{M} + y_{M}) + (a λ + μ) t^{'}$ so
$B (a x + y) = ψ (π (a x + y)) = ψ ((a λ + μ) t^{'}) = a λ + μ$
and $a B (x) + B (y) = a ψ (π (x)) + ψ (π (y)) = a ψ (λ t^{'}) + ψ (μ t^{'}) = a λ + μ$
By definition of $π, M \subseteq k e r π thus M \subseteq k e r B$ . Now, if $k e r B \neq M then k e r B$ is a subspace of $R^{p}$ containing strictly a $(p - 1)$ -subspace, hence $k e r B$ would be all $R^{p}$ . This is false because $B t^{'} = 1$ . Thus $k e r B = M$ .

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