# How many projections $\pi: \mathbb{F}_3^2 \to \mathbb{F}_3^2$ there? [closed]

How many projections $$\pi: \mathbb{F}_3^2 \to \mathbb{F}_3^2$$ there? Justify your answer!

If someone can give me a tipps on this question.D

• What is $F_3^2$? I'm thinking of this as $\mathbb{F}_3^2$ but not sure what that has to do with tensor products. Jul 9 '20 at 17:20
• @Osama Ghani it actually $\mathbb{F}_3^2$ Jul 9 '20 at 17:36
• What does this have to do with tensors? Jul 9 '20 at 17:56
• @Lubin please can you explain it cleary what is meant by the cardinality of the general linear group $\text{GL}^2(\Bbb F_3)$.I learn in german and it not easy getting the technical word so easily in english.thanks Jul 9 '20 at 20:24
• “Cardinality” means the number of things in the set. I’ve deleted the comment that gave rise to your request, because I see from @OliverClarke’s response that I seriously misinterpreted your question. Jul 9 '20 at 20:58

For clarity here's a definition of projection. Given a vectorspace $$V$$, a projection is a linear map $$\pi : V \rightarrow V$$ such that $$\pi \circ \pi = \pi$$. In particular $$\pi$$ acts as the identity on its image $$Im(\pi) \subseteq V$$ so we let's take cases on the rank of $$\pi$$.
I'll do one of the cases and leave the rest for you. Suppose that $$\pi$$ has rank one. First we count the number of possible images for each projection, i.e. the number of $$1$$-dimensional subspaces of $$\mathbb F_3^2$$. This is equal to the number of non-zero vectors up to scalar which is $$(9 - 1)/2 = 4$$ i.e. $$9-1$$ non-zero vectors and $$2$$ non-zero scalars in $$\mathbb F_3$$.
Now we need to count the number of possible projections for each possible image. Take a basis for the image of a projection and extend it to a basis for the whole space. In this case our projection has rank one so our basis has one extra non-zero vector. The projection must send this basis vector inside the $$1$$-dimensional subspace and any vector is possible. Since the subspace has dimension one, there are $$3$$ possible choices for the image of this vector.
So there are $$4 \cdot 3 = 12$$ projections of rank one. How many projections are there of rank zero and two?
• with rank zero, we are going to have no projektion,because we got no image,to extend to the rest of the field. when the rank=2.we still have 3 possibilities.Since we have already 2 non zero vektor.And when we extend it to the rest of the space.we obtained 3 possibility.that is $4.3=12$ projektion(rank=2) and 0 projektion (rank=0) Jul 9 '20 at 21:33