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Invariant subspaces of representations

Carla1985

Member
Feb 14, 2013
93
Let $\varrho :\mathbb{Z}\rightarrow GL_3(\mathbb{R})$ be the representation given by $\varrho (n)=A^n$ where

A=$\begin{pmatrix}
2 & 5 & -1 \\
2 & \frac{5}{2} & \frac{11}{2} \\
6 & \frac{-2}{2} & \frac{3}{2} \\
\end{pmatrix}$

Does ρ have any 1-dimensional invariant subspaces?


Do I have to divide up the elements in the representation with their corresponding eigenvectors. I know From $A^4$ the matrix is just a scaler of the previous 4, i.e. $A^4=I$, $A^5=A$ etc and their eigenvectors are the same then?

Im struggling to understand this chapter we are doing a little so any help is much appreciated.

Carla x
 
Last edited by a moderator:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Let $\varrho :\mathbb{Z}\rightarrow GL_3(\mathbb{R})$ be the representation given by $\varrho (n)=A^n$ where

A=$\begin{pmatrix}
2 & 5 & -1 \\
2 & \frac{5}{2} & \frac{11}{2} \\
6 & \frac{-2}{2} & \frac{3}{2} \\
\end{pmatrix}$

Does ρ have any 1-dimensional invariant subspaces?


Do I have to divide up the elements in the representation with their corresponding eigenvectors. I know From $A^4$ the matrix is just a scaler of the previous 4, i.e. $A^4=I$, $A^5=A$ etc and their eigenvectors are the same then?

Im struggling to understand this chapter we are doing a little so any help is much appreciated.

Carla x
Hey Carla! :)


Here's a definition from http://www.win.tue.nl/~aeb/2WF02/fgreps.pdf:

Let G be a finite group.
A linear representation of G is a homomorphism ρ : G → GL(V ) where GL(V ) is the group of invertable linear transformations of the vector space V.
A subspace W of V is called ρ(G)-invariant if ρ(g)W ⊆ W for all g ∈ G.


A 1-dimensional subspace of $\mathbb R^3$ would be the span of some vector $v$.
So $W = \{\mu v : \mu \in \mathbb R \}$.

What we need is that $\varrho(n)W ⊆ W$ for all $n \in \mathbb Z$.
What does this expand to for, say, $n=1$?

ILS x
 
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ThePerfectHacker

Well-known member
Jan 26, 2012
236
Let $\varrho :\mathbb{Z}\rightarrow GL_3(\mathbb{R})$ be the representation given by $\varrho (n)=A^n$ where

A=$\begin{pmatrix}
2 & 5 & -1 \\
2 & \frac{5}{2} & \frac{11}{2} \\
6 & \frac{-2}{2} & \frac{3}{2} \\
\end{pmatrix}$

Does ρ have any 1-dimensional invariant subspaces?


Do I have to divide up the elements in the representation with their corresponding eigenvectors. I know From $A^4$ the matrix is just a scaler of the previous 4, i.e. $A^4=I$, $A^5=A$ etc and their eigenvectors are the same then?

Im struggling to understand this chapter we are doing a little so any help is much appreciated.

Carla x
Let $G = \mathbb{Z}$ be additive group.
Let $V = \mathbb{R}^3$ be real vector space.
With linear action of $G$ on $V$ given by,
$$ n\cdot x = A^n x \text{ for }n \in G, x\in V$$

Say $W$ is a one-dimensional subspace of $V$, so $W = \left< w \right>$ is spanned by a non-zero vector in $\mathbb{R}^3$. Since $W$ is invariant under $G$ it means in particular that $1\cdot w \in W$, in other words, $Aw = kw$ for some $k\in \mathbb{R}$.

This shows that an invariant one-dimensional subspace of $V$ must be spanned by an eigenvector of $A$. Now look for the eigenvectors.
 
Last edited by a moderator:

Carla1985

Member
Feb 14, 2013
93
Hey Carla! :)


Here's a definition from http://www.win.tue.nl/~aeb/2WF02/fgreps.pdf:
Let G be a finite group.
A linear representation of G is a homomorphism ρ : G → GL(V ) where GL(V ) is the group of invertable linear transformations of the vector space V.
A subspace W of V is called ρ(G)-invariant if ρ(g)W ⊆ W for all g ∈ G.


A 1-dimensional subspace of $\mathbb R^3$ would be the span of some vector $v$.
So $W = \{\mu v : \mu \in \mathbb R \}$.

What we need is that $\varrho(n)W ⊆ W$ for all $n \in \mathbb Z$.
What does this expand to for, say, $n=1$?

ILS x
With n=1 I have $AW⊆W$

The only eigenvector for A is (1,1,1). Does this mean my span and the invariant subspace would be W ={(x,x,x)∈R3 |x∈R}?

There are elements of the representation that don't fall into this though (every 4th element), is that ok?

Thanks, I think I'm slowly starting to understand it, I know my brain is on a go slow not helped by the fact I was up at 5am with the kids this morning :p
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
With n=1 I have $AW⊆W$
More specifically $\varrho(1)W = \{\varrho(1)\mu v : \mu \in \mathbb R \}
= \{\mu A v : \mu \in \mathbb R \}$.
This is only contained in $W$ iff $v$ is an eigenvector of $A$.


The only eigenvector for A is (1,1,1). Does this mean my span and the invariant subspace would be W ={(x,x,x)∈R3 |x∈R}?
That would be the only candidate then.


There are elements of the representation that don't fall into this though (every 4th element), is that ok?
I assume you are referring to $\varrho(4) = A^4$, which is apparently equal to $I$?
If so, is $I(1,1,1) \in W = \{\mu (1,1,1) : \mu \in \mathbb R \}$?
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
The only eigenvector for A is (1,1,1). Does this mean my span and the invariant subspace would be W ={(x,x,x)∈R3 |x∈R}?


I did not compute the eigenvectors of A but assuming it is (1,1,1) then you are correct that W = <(1,1,1)> which is exactly how you described it.

There are elements of the representation that don't fall into this though (every 4th element), is that ok?


What do you mean by that?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I think there is a typo in your original matrix.

I believe the matrix actually under consideration must be:

\(\displaystyle A = \begin{bmatrix}2&5&-1\\-2&\frac{5}{2}&\frac{11}{2}\\6&-\frac{3}{2}&\frac{3}{2} \end{bmatrix}\)

which satisfies:

$A^4 - 1296I = 0$

(1,1,1) is indeed an eigenvalue of this matrix, and since $A$ satisfies the polynomial:

$x^4 - 1296 = (x^2 + 36)(x + 6)(x - 6)$

any eigenvalue must be a root of this polynomial (so we have just two possible eigenvalues, 6 and -6).

It turns out the characteristic polynomial is $(x - 6)(x^2 + 36)$, which rules out -6 as an eigenvalue. Solving for $E_6$ we find the space is indeed spanned by (1,1,1).

It is trivial that if:

$Av = \lambda v$ that:

$A^kv = \lambda^kv$

so any power of $A$ has the same eigenvectors as $A$.

Thus $\langle(1,1,1)\rangle = E_6$ is indeed a one-dimensional invariant subspace of $\rho$.
 
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Carla1985

Member
Feb 14, 2013
93
Hi, thanks all, I woke up this morning and it all started to make sense (eventually!). Here's how I have wrote it all out, hopefully its right lol:

We need to find a 1 dimensional subspace of $\mathbb{R}^3$, call this U. As its 1 dimensional subspace then any element of U is an eigenvector of $\varrho (n)$ with corresponding eigenvalue $\lambda_n$.

Since A has eigenvector (1,1,1) with eigenvalue 6 then (1,1,1) is our 1 dimensional basis element, call this u.

So $Au=6u$

And then:
$A^2u=A\lambda u=\lambda(Au)=\lambda(\lambda u)=\lambda^2u$

So then for any value of n:
$A^n u=\lambda^n u$

ie:
$\varrho (n)(u)=\lambda^n (u)\ \ \forall n \in \mathbb{N}$

Hence every element of U is a scaler multiple of u and U is a 1 dimensional invariant subspace of U.

Carla x
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Looks fine to me! :)

ILS x
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Looks good.