2 and 3 dimensional invariant subspaces of R4

  • #1
nigelscott
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I am looking at the representation of D4 in ℝ4 consisting of the eight 4 x 4 matrices acting on the 4 vertices of the square a ≡ 1, b ≡ 2, c ≡ 3 and d ≡ 4.

I have proven that the 1-dimensional subspace of D4 in ℝ2 has no proper invariant subspaces and therefore is reducible. I did this in 2 ways: Computing the null space of the 2 x 2 operators which turns out to be trivial and showing that the result of each operation cannot be a scalar multiplication of itself.

I now want to find the 2 and 3-dimensional subspaces of ℝ4 and determine whether they are proper invariant subspaces and are therefore reducible.

Attempt:

Writing as row vectors, I have found a basis of the null space of each as follows:

S = {(a b c d) : c = a + b, d = -2c} which gives {(1 1 0 0),(-1/2 0 1/2 1)}

and,

S = {(a b c d) : a + b + c = d} which gives {(-1 1 0 0),(-1 0 1 0),(-1 0 0 1)}

Is it correct to say that because both subspaces are non-trivial these representations are reducible? If not, what is the correct way to approach this problem?
 

Answers and Replies

  • #2
fresh_42
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It looks as if you confuse reducible and irreducible. E.g. one dimensional spaces are always irreducible, as they do not have any proper subspaces. A space is irreducible, if there is no proper invariant subspace, and reducible otherwise.

Next: What is invariant? Are you looking for spaces which are invariant under the entire group, or do you consider subgroups?

Could you list the eight group elements, as the dihedral groups have a few presentations. And how the operation is defined, too. Matrix multiplication?

Here is how to write matrices:
https://www.physicsforums.com/help/latexhelp/
 

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