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

Posted on January 26, 2009 by saij

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

Filed under: Algebra, Linear Algebra, Proof of the Day | Tagged: , , , , , ,

« »

Leave a Reply