$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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result
Concepts
If
Only If
proof
Let $V$, $W$ be vector spaces and $T:V\to W$ be linear. Then $T$ is injective if and only if $T$ maps linearly independent sets to linearly independent sets.
That is, if $S\subseteq V$ is linearly independent, then $T(S):=\{Tv:v\in S\}$ is linearly independent.

Concepts

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If

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Only If

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Proof

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Let $V$, $W$ be vector spaces and $T:V\to W$ be linear. Then $T$ is injective if and only if $T$ maps linearly independent sets to linearly independent sets.
That is, if $S\subseteq V$ is linearly independent, then $T(S):=\{Tv:v\in S\}$ is linearly independent.

Concepts

Coming soon

If

Coming soon

Only If

Coming soon

Proof

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FullPage
result
concepts
If
Only If
proof
FullPage
result
Concepts
If
Only If
proof
Let $V$, $W$ be vector spaces and $T:V\to W$ be linear. Then $T$ is injective if and only if $T$ maps linearly independent sets to linearly independent sets.
That is, if $S\subseteq V$ is linearly independent, then $T(S):=\{Tv:v\in S\}$ is linearly independent.

Concepts

Coming soon

If

Coming soon

Only If

Coming soon

Proof

Coming soon
Let $V$, $W$ be vector spaces and $T:V\to W$ be linear. Then $T$ is injective if and only if $T$ maps linearly independent sets to linearly independent sets.
That is, if $S\subseteq V$ is linearly independent, then $T(S):=\{Tv:v\in S\}$ is linearly independent.

Concepts

Coming soon

If

Coming soon

Only If

Coming soon

Proof

Coming soon
FullPage
result
concepts
If
Only If
proof