Most important basis vector?

Question

My question is this: Is there any notion of importance of a basis vector? In other words, when describing a vector space using a basis, are all basis vectors equally important or is there any notion or mathematical idea of ranking basis vectors? This is something similar to how the singular vectors in SVD can be ranked.

Answer 1

One of the important idea of undergrad linear algebra is that of finding a basis for a vector space $V$ such that a given a linear operator $T: V\to V$ can be easier to understand (e.g. decomposing $T$, express $T$ in a simpler form, diagonalizing $T$,...) In that sense, the answer to your question is yes.

Answer 2

I think what you are really after is the following:

Let's say one has a matrix $ A $ that represents some linear transformation. Then if you computer $ y = A x $, both $ x $ and $ y $ are expressed in the standard basis.

If $ A $ is diagonalizable, then you know that $ D = X^{-1} A X $ for diagonal $ D $. The columns of $ X $ are then eigenvectors of $ A $.

Now, let's look at $ y = A x $. You can rewrite this as $ X ( X^{-1} y ) = A X X^{-1} $ or $ \widehat y = X^{-1} A X \widehat x $ where $ \widehat y = X^{-1} y $ and $ \widehat x = X^{-1} x $. But that meants that $ \widehat y = D \widehat x $.

What do you notice:

• $\widehat y$ tells you the coefficients relative to the basis that form the columns of $ X $.
• $\widehat x$ tells you the coefficients relative to the basis that form the columns of $ X $.
• If you view the two vectors in that basis, the transformation $ y = A x $ requires the multiplication with a diagonal matrix $ D $ instead. Obviously, much simple.

The point: if you view your problem in the right basis, the transformation becomes much simpler. Usually this means you can draw conclusions more easily.