# When are linear transformations not invariant?

1. Jun 6, 2009

### evilpostingmong

I am studying invariance, and I came across this dilemma.
Suppose we have a subspace with the basis <v1, v2> of the subspace (lets say U2)
and we were to map v=c1v1+c2v2 and we let c2=0.
Now c1T(v1)+c2T(v2)=k1c1v1+0*T(v2)= k1c1v1.
I am doing a proof and need to
know what the question means by the intersection of a collection of
subspaces, and I believe that this is what it refers to,
since we can map c1v1 of <v1> (the basis of "U1") to U1 and arrive at the same

Last edited: Jun 6, 2009
2. Jun 6, 2009

### HallsofIvy

Staff Emeritus
First, it is not the linear transformation that is or is not invariant, it is a set of vectors that is or is not invariant under a transformation.

So T(c1v1+ c2v2)= T(c1v1) (because c2=0) T(c1v1)= c1T(v1). Now where do you get that c1T(v1)= k1c1v1? That is equivalent, of course, to saying that T(v1)= k v1 so that v1 is an eigenvector of T with eigenvalue k. The "intersection" of a collection of subspaces is just the intersection of the sets: those vectors that are in all of the subspaces. In linear algebra it is comparitively easy to show that the intersection of a collection of subspaces is itself a subspace.

I cannot see that "intersection of a collection of subspaces" has anything to do with T(c1v1). What, exactly, are you trying to prove?

3. Jun 6, 2009

### evilpostingmong

Oh sorry, kind of got caught up in the definition of invariance.
I do know how to prove this, but it was the definition that
got me stuck, but you made it a bit more clear. Thank you!