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Proof of the Day: The Kernel’s Injective Fried Chicken

Let $T : V\longrightarrow W$ be a linear map between vector spaces. Show that T is injective if and only if $ker T = \{0\}$.

Proof:

Suppose first that T is injective. Then $Tx = 0 = T0$ implies $x = 0$, so $ker(T) = \{ 0\}$.

Now suppose $ker(T) = \{ 0\}$. Then $Tx = Ty$ implies $T(x - y) = 0$ which implies $x - y = 0$, and $x = y$. So, T is injective.

$\Box$

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