Often it is said that one of the most useful properties of eigenvalues of a matrix is that they are invariant under change of basis. This in turn is said to be useful in physics because real, physical quantities represented by the matrix/tensor do not care about the coordinate system used to measure them. Therefore what we measure in real life needs to be independent of the coordinate system. Components of a tensor/matrix change with a change of basis, so we need to use the invariants of the tensor/matrix if the want to get correct results. This is what I think is true.

I'm having a hard time understanding this. I'm studying structural mechanics at the moment, where we write stresses in form of a tensor. The 3x3 stress tensor represents stresses at each point inside an object with normal stresses in diagonal and shear stresses off diagonal. Each column of the stress tensor gives a vector of stresses for each side of an infinitesimal "box" element. But the components of this tensor change with a change in coordinate system. Why is this a problem?

Let's say a particle has the velocity vector $(1, 1)$ in regular orthogonal cartesian coordinates. If we introduce another rectangular coordinate system that is rotated 90 degrees counter-clockwise, the particle now has the velocity vector $(1, -1)$. Even though the vector has different components, neither vector is wrong: They both represent the same velocity on real life.

So why we need to resort to using invariants of a tensor to calculate things in real life if the components of the tensor do represent the same thing, but just using different coordinate system? Thank you!


For example Wikipedia page on Cauchy stress tensor states:

The components ${\displaystyle \sigma _{ij}} $ of the stress tensor depend on the orientation of the coordinate system at the point under consideration. However, the stress tensor itself is a physical quantity and as such, it is independent of the coordinate system chosen to represent it.


The value of these components will depend on the coordinate system chosen to represent the vector, but the magnitude of the vector is a physical quantity (a scalar) and is independent of the Cartesian coordinate system chosen to represent the vector.

In addition, I got an answer on my earlier question on Engineering SE stating that independence from coordinate system is necessary, or any conclusions drawn from the analysis are wrong.


A vector changes components if you change coordinate system but it remains the same vector, the same "arrow". If A points a finger (a vector) at B, it will be pointing at B in every coordinate system. In the case of velocity, yes, the vector will change components but the speed (the norm, the invariant) will be the same. Observers in different frames of reference are describing the particle moving with different components but for both of them the speed has to be the same (ignore relativity here, ofc)

Scalar quantities (energy, principal stress) can not depend on the choice of coordinate system. Indeed, the eigenvectors WILL change components but the eigenvalues will stay the same. Scalars refer to measurable quantities, indipendent from the frame of reference. You cannot "measure" a vector, you can only describe it with components. Vectors are only representations, but whatever description you choose it will have to be compatible with some comnstraint (e.g. the norm can not change). The fact that me and you disagree on the singular component is meaningless, the important fact is that we agree on the global representation which is the physical one (e.g. we get the same speed or A is pointing the finger at B).

Part of a physical problem will be described in a subjective way (which frame of reference one chooses for example) but the physical quantities can not vary. In the case of vectors (tensors), the physical quantity is the vector (tensor) per se, as a "metaphysical" object, as an "arrow". That also stays the same, whatever coordinate system we choose. We are just describing it in different ways, as if changing unit of measure (mph rather than m/s for speed).


This claim is just wrong. Since you haven't told us any context, the name of the author, or an accurate quote, it's hard to evaluate this in detail. Maybe you're misunderstanding what they said, or they're saying something more nuanced that you're missing.

As a counterexample from general relativity, see Can GR be reformulated in terms of invariant observables? All curvature invariants vanish for any gravitational plane-wave solution in 3+1 dimensions. This is basically because an observer who chases the wave at high velocity can see the energy and amplitude of the wave Doppler shifted down to an arbitrarily low level.

  • $\begingroup$ Please see the edits on my question. $\endgroup$ – S. Rotos Jun 8 at 22:56

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