Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $x$ be a variable. Denote by $V$ the vector space consisting of all polynomials $P(x)=ax^2+bx+c$ of degree not more than 2, with complex coefficients. For any real number $t$ determine an operator $\varphi(t):P \mapsto \tilde{P}$ on the vector space $V$ by the formula $\tilde{P}(x) = P(x+t)$. Consider the set $\mathbb{R}$ of real numbers as a group with the additive composition law.

(a) Show that the correspondence $t \mapsto \varphi(t)$ is a representation of the group $\mathbb{R}$ in the complex vector space $V$.

(b) Demonstrate that the representation $\varphi$ is not irreducible.

(c) Describe all vector subspaces of $V$ that are preserved by every operator $\varphi(t)$ where $t$ ranges over $\mathbb{R}$. Prove that the representation $\varphi$ is not decomposable into a direct sum of irreducible representations of the group $\mathbb{R}$.

For part (c), we know that the non-trivial $\varphi$-invariant subspaces of $V$ will be 1 or 2-dimensional. I've found the possible 1-dimensional $\varphi$-invariant subspaces of $V$ by solving the equation $$\varphi(t)(ax^2+bx+c)=\alpha(ax^2+bx+c)$$ for $a,b,c$, where $\alpha$ is a scalar. We get that the only 1-dimensional $\varphi$-invariant subspace of $V$ is that consisting of constant polynomials.

How would I find the possible 2-dimensional $\varphi$-invariant subspaces, and hence show that $V$ can't be expressed as a direct sum of a 1-dimensional and a 2-dimensional invariant subspace (and so is not decomposable)?

share|improve this question
Wherever did you get that formatting from? –  Chris Eagle Nov 23 '12 at 17:34
Sorry, just getting used to the Tex –  Sam Nov 23 '12 at 17:36
should be readable now! –  Sam Nov 23 '12 at 17:40
typed out the full question now –  Sam Nov 23 '12 at 17:53

2 Answers 2

Remember that invariant subspaces of a representation $\pi:G\to GL(V)$ are the linear spans of sets of the form $\{ \pi(g) v\vert g\in G, v\in A\}$ for $A\subseteq V$. So a good starting point is studying the spans of orbits of points.

  • You can check directly that the orbits of nonzero constant polynomials are just singletons, and they span the space of all constant polynomials;
  • the orbits of linear polynomials are sets of the form $V'_a=\{\alpha(x+a+t)\vert t\in {\bf R}\}$ for some $a,\alpha\in {\bf C}$ with $\alpha\neq 0$, and it is easy to see that they span a space containing constants and from that it's easy to see that they all span the entire space of polynomials of degree $\leq 1$;
  • similarly, you can show that the span of each orbit of a quadratic contains a linear polynomial and hence its orbit, so from that, all linear polynomials, and so also all quadratics, so it is the entire space.
share|improve this answer
So what does this tell us about what the 2-dimensional $\varphi$-invariant subspaces are? –  Sam Nov 23 '12 at 18:39
The only $\varphi$-invariant 2-dimensional subspace is the one consisting of all polynomials of degree at most 1? –  Sam Nov 23 '12 at 19:01
@Sam: yes, because any invariant subspace contains the orbit of some point, and there is only one $1$-dimensional invariant subspace. –  tomasz Nov 23 '12 at 19:16
So $\varphi$ is not decomposable since the only possible decomposition into non-trivial invariant subspaces would be $V=W \oplus W'$ where $$W= \{ \text{polynomials of degree at most 1} \}$$ $$W'=\{\text{constant polynomials} \}$$ but $x^2 \in V$ and $x^2 \neq (bx + c) + d$ for any $bx+c \in W, d\in W'$ so that $V \neq W \oplus W'$. –  Sam Nov 23 '12 at 19:29
@Sam: or you could just notice that $W'\subseteq W$. ;) –  tomasz Nov 23 '12 at 21:17

Here is a possible way: a subspace $U$ of space $V$ is preserved by operator $\varphi \colon V \to V $ iff its annihilator $U^0$, which is a subspace of the dual space $V^*$, is preserved by the dual operator $\varphi^* \colon V^* \to V^*$. Plus, if $\dim U = 2$ and $\dim V = 3$, then $\dim U^0 = 1$. So it comes down to finding $1$-dimensional invariant subspaces of $V^*$, which can be done by finding the eigenvectors of $\varphi^*$.

share|improve this answer
I'm not familiar with dual spaces. Is there another way? –  Sam Nov 23 '12 at 18:25
@Sam Then tomasz's answer suits you better than mine. This is actually how I originally solved this. It just seemed that writing a dual space argument would be less work for my fingers )) –  Dan Shved Nov 23 '12 at 18:42
@Sam Plus, tomasz's approach can be easily generalized to higher dimensions, and mine cannot. –  Dan Shved Nov 23 '12 at 18:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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