What does $QAQ^{-1}$ actually mean

coordinate systemslinear algebralinear-transformationsmatricesvector-spaces

I'm self-learning Linear Algebra and have been trying to take a geometric approach to understand what matrices mean visually. I've noticed this matrix product pop up repeatedly and can't seem to decipher what it means.

Let me provide some context. I was specifically looking at the application of linear algebra to the stress tensor $\sigma$ and how we get the transformed stress state in some rotated coordinate system by the following multiplication:$\sigma'=Q\sigma Q^{-1}(Q$ is a rotation matrix defined by the angle we want to rotate the axes).

I might be missing something key here but I'm unsure as to why we need this product at all. From what I know, if we have a rotation matrix, left multiplying some vector by this is akin to transforming the basis and that very vector with the axes fixed. Equally, we can think of this from a different perspective as rotating the axes the opposite direction with the vector fixed.

I guess my question is then, why do we not just compute $\sigma'=Q \sigma$ to get the rotated coordinates for stress?

I hope I was clear, thanks in advance.

Best Answer

Let's say we have the matrix acting on an arbitrary vector

$$w = Av$$

where $v\mapsto Av$ is a linear transformation $T:\Bbb{R}^n \to \Bbb{R}^n$. If we want to "rotate" the matrix, we have to check and see what kinds of rules such a transformation to follow. The relationship we want to hold is

$$w' = A'v'$$

Then plugging in $w'=Qw$ and $v'=Qv$ we get that

$$Qw = A'Qv \implies QAv = A'Qv $$

for all $v$, thus

$$A' = QAQ^{-1}$$

Related Question