Are linear automorphisms nothing but the identity mapping?

Hi Physics Forums,
Quick question! Are every automorphism on a vectors space ## V ## over some field ## \mathbb{F} ## nothing but the identity mapping in disguise?

The reason for asking is; automorphisms are (from my point of view) basically a change of basis, and vectors are invariant under change of basis (their coordinates do indeed change)... So automorphisms are taking every vector to itself?

Thnx

William Crawford said:

TL;DR Summary: Are linear automorphisms nothing but the identity mapping?

Hi Physics Forums,
Quick question! Are every automorphism on a vectors space ## V ## over some field ## \mathbb{F} ## nothing but the identity mapping in disguise?

The reason for asking is; automorphisms are (from my point of view) basically a change of basis, and vectors are invariant under change of basis (their coordinates do indeed change)... So automorphisms are taking every vector to itself?

Thnx

No. A rotation, say at a right angle, in the plane doesn't fix anything other than the zero vector.

The flaw of your argument lies in the consideration of your bases. You can have a vector ##(1,0)## in base ##1,## say ##(1,0) \triangleq 1\cdot b_1^{(1)}+ 0\cdot b_2^{(1)}.## Now we apply an automorphism ##A## and get ##A(1,0)\triangleq (a_1,a_2)=1\cdot Ab_1^{(1)}+0\cdot Ab_2^{(1)}.## The vectors ##(a_1,a_2)## and ##(1,0)## are not equal in general, i.e. for example if ##A## is the rotation @martinbn mentioned.

Now comes your magic trick: you define ##b_1^{(2)}:=Ab_1^{(1)}## and ##b_2^{(2)}:=Ab_2^{(1)}## which makes your new vector ##(a_1,a_2)=1\cdot b_1^{(2)}+0\cdot b_2^{(2)} \triangleq (1,0).## Now your new vector looks like the old one, however, you changed your yardstick! You are considering a different basis ##2##. This only proves that an automorphism that changes vectors can be described in a new basis as if there was no change, using the same coordinates. But then you have transformed the coordinate system, the basis. So if you first apply an automorphism and then a certain change of basis, then it may look like you haven't changed anything because both effects cancel out.

If you only apply an automorphism, then the coordinates of your vector will change unless it is the identity. If you only apply a change of basis (a different yardstick), then the coordinates of your vector change unless it was the identity. If you do both in a certain way, both effects can cancel out.

William Crawford said:

TL;DR Summary: Are linear automorphisms nothing but the identity mapping?

Hi Physics Forums,
Quick question! Are every automorphism on a vectors space ## V ## over some field ## \mathbb{F} ## nothing but the identity mapping in disguise?

The reason for asking is; automorphisms are (from my point of view) basically a change of basis, and vectors are invariant under change of basis (their coordinates do indeed change)... So automorphisms are taking every vector to itself?

Thnx

You should be aware that the abstract definition of a vector in mathematics is very different from the physics definition of a vector. In mathematics, a vector is a member of a vector space, which has simple requirements. Being invariant under a change of basis is not part of the abstract mathematical definition.

William Crawford said:

automorphisms are (from my point of view) basically a change of basis

This part is true. Isomorphisms must map bases to bases. But it does not imply an automorphism is the identity.

martinbn said:

No. A rotation, say at a right angle, in the plane doesn't fix anything other than the zero vector.

Maybe even simpler is x -> 2x from ##\mathbb{R}## to ##\mathbb{R}##.

@William Crawford: I believe you are thinking of a change of coordinates. Coordinates are closely related to bases. I.e. a basis for V defines, and is defined by, an isomorphism A:R^n-->V, (namely the A-basis is the image of the standard one, (1,0,...,0), (0,1,0,...).....)), and whose inverse A^-1:V-->R^n is the corresponding coordinate system (i.e. the A- coordinates of v are the entries of the vector A^-1(v) in R^n.)

If B:R^n-->V, is determined by another basis, we can change from the A coordinates to the B coordinates by the composition M = (B^-1.A):R^n-->V-->R^n, an automorphism, not of V, but of coordinate space R^n. I.e.a change of coordinates for V, is given by an automorphism of R^n, not of V.

The corresponding non - trivial automorphism of V, B.A^-1:V-->V maps the A basis into the B basis. This transformation of V that maps one basis into another, which is what some may be thinking of as a change of basis map, is obviously not the identity.

This is all a bit confusing of course, since if M = B^-1.A is the change of coordinate matrix, then the map B.M.A^-1:V-->V, which assigns to v its A coordinates, then changes those coordinates to the B coordinates of v, then goes back to the vector with those B-coordinates, namely v, is indeed the identity map. This I think is what you meant by your original remark.

I hope this helps somebody. But this may be one of those infinite sinks of explanation energy, like why .9999.... = 1.