# 2 and 3 dimensional invariant subspaces of R4

• I
• nigelscott
In summary, the conversation discussed the representation of D4 in ℝ4 and its 1-dimensional subspace in ℝ2. It was proven that the subspace is reducible, and the conversation then moved on to finding the 2 and 3-dimensional subspaces of ℝ4 and determining if they are proper invariant subspaces. Two bases for the null space of each subspace were found and it was questioned if this means the representations are reducible. It was clarified that one-dimensional spaces are always irreducible and that the correct way to approach the problem is to determine if there are proper invariant subspaces. The group elements and operation were also requested for clarification. Matrices were provided as a guide for writing them in future discussions.
nigelscott
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?

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/

## 1. What are invariant subspaces in R4?

Invariant subspaces in R4 are subspaces that remain unchanged under a linear transformation. This means that the vectors within the subspace are mapped to themselves by the transformation.

## 2. How many 2-dimensional invariant subspaces are there in R4?

There are infinitely many 2-dimensional invariant subspaces in R4. This is because any two linearly independent vectors in R4 can form a 2-dimensional subspace that is invariant under certain linear transformations.

## 3. Can a 3-dimensional subspace be invariant in R4?

Yes, a 3-dimensional subspace can be invariant in R4. This is because any three linearly independent vectors in R4 can form a 3-dimensional subspace that is invariant under certain linear transformations.

## 4. How do you determine if a subspace is invariant in R4?

To determine if a subspace is invariant in R4, you can apply the linear transformation to the basis vectors of the subspace. If the resulting vectors are still within the subspace, then the subspace is invariant.

## 5. Can an invariant subspace in R4 have a basis that is not orthogonal?

Yes, an invariant subspace in R4 can have a basis that is not orthogonal. However, having an orthogonal basis can make it easier to determine if a subspace is invariant, as the basis vectors will remain orthogonal under a linear transformation.

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