Question

Let $(X, || \cdot ||_x)$ and $(X, || \cdot ||_y)$ be two normed spaces. We identify $X$ with $Y$ as normed spaces (that is, they are indistinguishable) when there is a linear map $T: X \to Y$ such that $T$ is (i) one-to-one, (ii) onto and (iii) isometry or norm preserving: $||Tx||_y = ||x||_x$. We call $T$ an equivalence or congruence. Consider the linear map $S: X \to Y$. (a) Prove that $S$ is a congruence $\iff$ $S$ is onto and is an isometry. (b) Prove that $S \circ T$ and $T^{-1}$ is a equivalence if $S, T$ is an equivalence.

          Let $(X, || \cdot ||_x)$ and $(X, || \cdot ||_y)$ be two normed spaces. We identify $X$ with $Y$ as normed spaces (that is, they are indistinguishable) when there is a linear map $T: X \to Y$ such that $T$ is (i) one-to-one, (ii) onto and (iii) isometry or norm preserving:

$||Tx||_y = ||x||_x$.

We call $T$ an equivalence or congruence.

Consider the linear map $S: X \to Y$.

(a) Prove that $S$ is a congruence $\iff$ $S$ is onto and is an isometry.

(b) Prove that $S \circ T$ and $T^{-1}$ is a equivalence if $S, T$ is an equivalence.

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Let (X, || · ||x) and (X, || · ||y) be two normed spaces. We identify X with Y as normed spaces (that is, they are indistinguishable) when there is a linear map T: X → Y such that T is (i) one-to-one, (ii) onto and (iii) isometry or norm preserving:

||Tx||y = ||x||x.

We call T an equivalence or congruence.

Consider the linear map S: X → Y.

(a) Prove that S is a congruence S is onto and is an isometry.

(b) Prove that S ∘ T and T^-1 is a equivalence if S, T is an equivalence.

Added by Charles B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/27/2023

Step 1

Step 1: To prove that S is a congruence if and only if S is onto and is an isometry, we need to show both directions of the implication. Show more…

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Let (x,||*||_(x)) and (x,||*||_(Y)) be two normed spaces. We identify x with Y as normed spaces (that is, they are indistinguishable) when there is a linear map T:x->Y such that T is (i) one-to-one, (ii) onto and (iii) isometry or norm preserving: ||Tx||_(Y)=||x||_(x). We call T an equivalence or congruence. Consider the linear map S:x->Y. (a) Prove that S is a congruence <=>S is onto and is an isometry. (b) Prove that S@T and T^(-1) is a equivalence if S,T is an equivalence. Let (X,|[x) and (X,|[y) be two normed spaces. We identify X with Y as normed spaces (that is , they are indistinguishable) when there is a linear map T : X - Y such that T is (i) one-to-one, (ii) onto and (iii) isometry or norm preserving: |Tx|y=|x|x We call T an equivalence or congruence. Consider the linear map S : X - Y. a Prove that S is a congruence S is onto and is an isometry. (b) Prove that S o T and T-1 is a equivalence if S,T is an equivalence

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