Injective linear operator question

Zermelo

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Jan 7, 2021
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Hello there, my Linear Algebra textbook says an interesting thing: "A linear operator T: V ->W is injective if and only if there doesn't exist any element x from V, \(\displaystyle x \neq 0v\) such that T(x) = 0w".
While the "if", aka "If T is injective, there doesnt exist..." makes sense and is pretty trivial, but I don't see any reason for the "only if".
Let's suppose that an element \(\displaystyle x \neq 0v\) such that T(x) = 0w doesn't exist. Why does this imply that the operator is injective? I really don't see a problem with having some two arbitrary elements x and y from V such that \(\displaystyle x \neq y\) and T(x) = T(y) under these circumstances. Am I missing something or did my professor make a mistake?
NOTE: 0v and 0w are the neutral elements of vector spaces V and W, respectfully.
 

Jomo

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Dec 30, 2014
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Fact: T: V ->W is injective if and only if there doesn't exist any element x from V, x≠0v such that T(x) = 0w".

Proof: (<=)Suppose T(a) = T(b). Then T(a) - T(b) = 0w. So T(a-b) = 0w. Since only T(0v) = 0w, it follows that a-b = 0v or a=b. But if a=b, then T is 1-1.


You do realize that there doesn't exist any element x from V, x≠0v such that T(x) = 0w means that only T(0)=0w?
 

Zermelo

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Jan 7, 2021
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Fact: T: V ->W is injective if and only if there doesn't exist any element x from V, x≠0v such that T(x) = 0w".

Proof: (<=)Suppose T(a) = T(b). Then T(a) - T(b) = 0w. So T(a-b) = 0w. Since only T(0v) = 0w, it follows that a-b = 0v or a=b. But if a=b, then T is 1-1.


You do realize that there doesn't exist any element x from V, x≠0v such that T(x) = 0w means that only T(0)=0w?
Nice! Now I do realize. Thanks a lot!
 
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