# What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?

What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?

I know from the answers at the back of the book that the following map:

$$(S(t)f)(x) = f (tx),$$ where $$S: R \rightarrow S(V)$$ and $$V$$ is the subspace of all polynomials with real coefficients and $$t \in \mathbb{R}, f \in V.$$

Is not a linear representation (but I do not know how?). Even though I found it is linear when I put instead of $$f$$, $$\alpha f + \beta f$$. I know that for the given map to be a linear representation it must be a homomorphism (I am not very sure from this information, is it correct?) so what are the operations that I should consider here for studying homomorphism?

Could anyone clarify this discrepancies for me please?

• What is $R$ in this case? What is $S(V)$? – Omnomnomnom Jan 12 at 17:19
• R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom – hopefully Jan 12 at 17:25
• Which book are you using? – user458276 Jan 12 at 17:36
• @user458276 "Linear Representations of groups" for Ernest B. Vinberg – hopefully Jan 12 at 17:56

For all $$t$$, $$S(t)$$ is indeed a linear map; but $$t\mapsto S(t)$$ is not a morphism, because $$S(t+t') \neq S(t)\circ S(t')$$ in general.

If you consider $$\mathbb{R}^*\to GL(\mathbb{R}[X])$$, $$t\mapsto S(t)$$, it will, however be a morphism because $$(S(tt')f )(x) = f(tt'x) = f(t\cdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)\circ S(t') (f))(x)$$ so $$S(tt') = S(t)\circ S(t')$$.

A linear representation of a group $$G$$ is a morphism $$G\to GL(V)$$, that is, for each $$g\in G$$ you have a linear map $$\rho_g : V\to V$$ (subject to certain conditions). $$\rho$$ is a group morphism, $$\rho_g$$ is linear.

Here $$S(t)$$ is linear but $$t\mapsto S(t)$$ is not a group morphism from $$(\mathbb{R},+)$$

• What is S(V) here and why $S(t + t^')$$\neq$$S(t) \circ S(t^')$, could you give me an example please? – hopefully Jan 12 at 18:01
• Why "$S(t+t') \neq S(t)\circ S(t')$ in general" , could you give me an example please? – hopefully Jan 12 at 18:08
• Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)\circ S(t')$. Pick almost any example of polynomial to see – Max Jan 12 at 18:17
• okay thank you so much :) – hopefully Jan 12 at 18:22
• It's probably the group of invertible linear operatirs (but not $n\times n$ because $V$ is infinite dimensional) – Max Jan 12 at 18:31