0
$\begingroup$

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

$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 9 '20 at 17:20
  • $\begingroup$ @Osama Ghani it actually $\mathbb{F}_3^2$ $\endgroup$
    – Nganja
    Jul 9 '20 at 17:36
  • $\begingroup$ What does this have to do with tensors? $\endgroup$ Jul 9 '20 at 17:56
  • $\begingroup$ @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 $\endgroup$
    – Nganja
    Jul 9 '20 at 20:24
  • $\begingroup$ “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. $\endgroup$
    – Lubin
    Jul 9 '20 at 20:58
1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Of course this is right. I’ll delete my inapposite remark. $\endgroup$
    – Lubin
    Jul 9 '20 at 20:59
  • $\begingroup$ 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) $\endgroup$
    – Nganja
    Jul 9 '20 at 21:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.