If is not injective, take , ; then are linearly independent and . This implies that ; in other words , so that if is not injective, there exists a two-dimensional subspace mapping to a one-dimensional subspace.
I got the following question wrong in the final exam : Tell whether the affirmation is true or false by justificating.
is an injective linear transformation if and only if for all subspaces of with dimension 2, is injective.
I said it was true but my explanation was wrong. I had a small doubt : I thought it might be false because can be uneven but then this doubt disappeared.
Can you help me with this exercise?
Not sure what you mean by "T has to be injective in order to...", but yeah I proved .
If is injective, then clearly the restriction of to any is also injective.
If is not injective, then there exists some 2-dimensional whose image is 1-dimensional; hence surely if all 2-dimensional have 2-dimensional images then must be injective.
This is a proof by the "contrapositive" - in order to prove that A implies B, we prove that (not B) implies (not A).