# L_A is isomorphism

1. Sep 23, 2012

### damabo

1. The problem statement, all variables and given/known data

if $L_A: ℝ^n -> ℝ^n : X-> A.X$ is a linear transformation, and A is an orthogonal matrix, show that L_A is an isomorphism.
also given is that $(ℝ,ℝ^n,+,[.,.])$ , the standard Euclidian space which has inproduct [X,Y]= X^T.Y

2. Relevant equations

ortogonal matrix, so $A^T=A^{-1}$
isomorphism = bijective and linear (so what is left to show is bijective)

3. The attempt at a solution

don't know where to start

2. Sep 23, 2012

### jbunniii

Well, what's the definition of bijective?

3. Sep 23, 2012

### damabo

surjective and injective ofcourse:
injective:
I must show that if $X_1 != X_2$ then $A.X_1 ≠A.X_2.$
So choose $X_1,X_2 \in ℝ^n$. because $[AX_1,AX_2]=X_1^T.A^T.A.X_2 = X_1^T.X_2=[X_1,X_2]$, length, distance and orthogonality will be preserved. so $A.X_1 ≠A.X_2$.
surjective:
I must show that for every $Y \in ℝ^n : A.X=Y$for some $X \in ℝ^n$
Choose Y in ℝ^n. Is there an X in ℝ^n such that A.X=Y? I'm not sure what to do here.

4. Sep 23, 2012

### Dick

You've shown A is injective, so you know the kernel of A is {0}. Use the rank-nullity theorem.