In article <1168177503.819256.189270@v33g2000cwv.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:

What can be said transformations of a traceless matrix?

I'm guessing a traceless matrix can always be transformed to a
coordinate system where it is purely off-diagonal. My motivation is
decomposing a deformation or stress system to a pure isotropic dilation
or hydrostatic stress, plus something else. It would be nice if the
"something else" turned out to be a linear combination of pure shear
stresses or strains in an appropriate coordinate system, but I can't
quite see it.

I assume you're talking about similarity transformations
A -> P A P^(-1) where P is an invertible matrix, and we're working
with n x n matrices over the real field.

The answer is yes.

Consider any A with trace(A) = 0. First look at the (1,1) entry.
If it is not 0, there is another diagonal entry, say the (i,i)
entry, with the opposite sign. Let S be the matrix with entries
S_{1,1} = S_{i,i} = cos(t), S_{i,1} = sin(t), S_{1,i} = -sin(t),
all other entries the same as the identity matrix. Then
for t = 0, S A S^(-1) = A, while for t = Pi/2, S A S^(-1)
interchanges the (1,1) and (i,i) entries. Therefore there is
some t for which (S A S^(-1))_{1,1} = 0.

Now suppose we have
[ W X ]
S A S^(-1) = [ Y Z ]
in block-matrix form, where all diagonal elements of W are 0,
and thus trace(Z) = 0.
For any (n-k) x (n-k) invertible matrix R, let
[ I 0 ] [ I 0 ]
T = [ 0 R ], so T^(-1) = [ 0 R^(-1) ]
[ W XR^(-1) ]
and note that T S A (T S)^(-1) = [ RY RZR^(-1) ]
As shown above, we can choose R so RZR^(-1) has its top left
entry 0, so T S A (T S)^(-1) has an additional 0 on its
diagonal. Repeating this at most n-1 times, we get a matrix
similar to A with all 0's on the diagonal.

In article <1168177503.819256.189270@v33g2000cwv.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:

What can be said transformations of a traceless matrix?

I'm guessing a traceless matrix can always be transformed to a
coordinate system where it is purely off-diagonal. My motivation is
decomposing a deformation or stress system to a pure isotropic dilation
or hydrostatic stress, plus something else. It would be nice if the
"something else" turned out to be a linear combination of pure shear
stresses or strains in an appropriate coordinate system, but I can't
quite see it.

I assume you're talking about similarity transformations
A -> P A P^(-1) where P is an invertible matrix, and we're working
with n x n matrices over the real field.

The answer is yes.

Consider any A with trace(A) = 0. First look at the (1,1) entry.
If it is not 0, there is another diagonal entry, say the (i,i)
entry, with the opposite sign. Let S be the matrix with entries
S_{1,1} = S_{i,i} = cos(t), S_{i,1} = sin(t), S_{1,i} = -sin(t),
all other entries the same as the identity matrix. Then
for t = 0, S A S^(-1) = A, while for t = Pi/2, S A S^(-1)
interchanges the (1,1) and (i,i) entries. Therefore there is
some t for which (S A S^(-1))_{1,1} = 0.

Now suppose we have
[ W X ]
S A S^(-1) = [ Y Z ]
in block-matrix form, where all diagonal elements of W are 0,
and thus trace(Z) = 0.
For any (n-k) x (n-k) invertible matrix R, let
[ I 0 ] [ I 0 ]
T = [ 0 R ], so T^(-1) = [ 0 R^(-1) ]
[ W XR^(-1) ]
and note that T S A (T S)^(-1) = [ RY RZR^(-1) ]
As shown above, we can choose R so RZR^(-1) has its top left
entry 0, so T S A (T S)^(-1) has an additional 0 on its
diagonal. Repeating this at most n-1 times, we get a matrix
similar to A with all 0's on the diagonal.

On Sun, 7 Jan 2007, Edward Green wrote:

What can be said transformations of a traceless matrix?

I'm guessing a traceless matrix can always be transformed to a
coordinate system where it is purely off-diagonal. My motivation is
decomposing a deformation or stress system to a pure isotropic dilation
or hydrostatic stress, plus something else. It would be nice if the
"something else" turned out to be a linear combination of pure shear
stresses or strains in an appropriate coordinate system, but I can't
quite see it.

Trace(A) is a scalar, in the sense of being invariant under rotations. So,
given a rotation R, trace(R A inv(R)) = trace(A). For a 3x3 matrix A, R
has 3 parameters. If we have A' = R A inv(A), demanding that A'11 = A'22
= A'33 = 0 would seem to add 3 constraints, implying a unique proper
rotation give off-diagonal A'. There would also be a matching improper
rotation. Well, that's some first thoughts; perhaps they are use to you.

Presumably there is some geometric meaning if A is a tensor. This question
is not unrelated to the "Jackson question" thread on sci.physics, circa
mid-December.