# Why invariance to change of basis is so important in linear algebra?

I'm reading a book on linear algebra and I see that for every new presented concept (from simple vectors and linear functions and up to tensors) we immediately study how does it behave under a change of basis. Is it invariant or not, etc. This idea seems to be extremely important, if not central for LA.

My question is - why? Why is it so important?

My only vague thought so far is that talking/thinking about something without use of matrices and numeric coordinate to represent it, can be simpler than with them. But this is the point where my understanding ends.

• because choice is a bad thing Dec 12, 2013 at 17:08
• Is the book written by physicists and/or engineers? Dec 12, 2013 at 17:37
• @rschwieb: no, "Linear Algebra" by Kostrikin, it's rather abstract Dec 12, 2013 at 17:43
• @izhak OK :) I was just curious. I was going to say that physicists and engineers seem to have their own ideas about linear algebra that my answer would probably not reflect well. Dec 12, 2013 at 17:56

Let's recall something briefly. A linear transformation of one vector space to another has an existence that doesn't depend on bases.

Bases are just "skeletons" of the vector space that we pick so that we can concretely write out the transformations in terms of matrices. Different choices of bases yield different matrices for a transformation, but the transformation itself does not require any of the bases. Only representations of the transformation require a basis.

We are interested in properties that all the representations share, because then we can rightly say it's a property of the transformation and not just a quirk about one particular representation. In a sense, when we pick a basis and represent a transformation with it, we have "zoomed in too far" and we might overlook information about "the big picture" (the transformation itself). That's why we study invariants, so we know what things in "the small picture" actually reflect the big picture.

So, for example, one thing we know is invariant between similar matrices is the determinant. This demonstrates that the determinant is a property of the transformation and not really the matrix. It can be computed from the matrix, but you'll get the same answer no matter what basis you picked.

• Thank you for your answer, it's great! If I understand you right, you say that properties which are invariant to change of basis are important to us since they are invariant to linear transformation of vector space. I had such thought, but two things confused me: first that change of basis was introduced in the book much earlier than linear transformations themselves (so this motivation became unclear) and, second, that linear transformations as I understand are not exactly equal to change of basis (e.g. rank(T) can be less than dim(V)). Dec 12, 2013 at 18:23
• @izhak The rank of a matrix is equal to the dimension of the image of the transformation T it represents denoted by $Im(T)$. You probably already know or will soon learn the rank-nullity theorem which says that $\dim(V)=\dim(ker(T))+\dim(Im(T))$. This naturally implies that the rank is always less than or equal to the dimension of $V$. The rank of a transformation is again not dependent on basis, so it's a property of the transformation. Good example! Dec 12, 2013 at 18:25

Certainly if some property that objects have is not dependent on the basis under which those objects are studied, that property is more readily identified and studied.