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.
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ Basically (pun not intended), you should never describe a vector space using a basis $\endgroup$– Hagen von EitzenMay 27, 2018 at 20:42
-
$\begingroup$ @HagenvonEitzen I think that people in a way do this when they construct finite elements. See e. g. Ciarlet definition of the finite element. But that is a bit far from the algebraic setting. $\endgroup$– KorfMay 27, 2018 at 21:27
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
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.
-
$\begingroup$ I wanted to know if there exists a criterion to describe the importance of basis vectors. Your answer seems more subjective. $\endgroup$ May 27, 2018 at 21:10
-
1$\begingroup$ The answer of @Chuong seems fine to me. A priori, all your basis have the same importance. That is, for an abstract vector space $V$ there isn't a predefinite basis. You know that there is at least one, and that is enough for you to prove many things. Now, if your vector space is endowed with an extra operasition the situation is different. For example, if you have an inner product, the basis for which the product is diagonal seems more natural or important or whatever than any other basis. The same happen if there is a linear transformation $T$ as he has said or any other structure. $\endgroup$– Dog_69May 27, 2018 at 21:21
-
$\begingroup$ Perhaps a rephrasing of my question would help? I meant that given a basis for a vector space, can we rank the basis vectors by, say, representation ability? Something along the lines of, say, the 'variance' of the space captured by the one dimensional subspace spanned by the vector. $\endgroup$ May 28, 2018 at 0:07
$\begingroup$
$\endgroup$
3
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.
answered May 28, 2018 at 6:56
-
$\begingroup$ Please read my comment to chuong's answer. $\endgroup$ May 28, 2018 at 11:06
-
$\begingroup$ I'm afraid I don't understand the comment. $\endgroup$ May 29, 2018 at 6:10
-
$\begingroup$ So here is an idea of how to quantify the importance of a basis. Take the basis, make it into the columns of $ X $. Then compute $ X^{-1} A X $ and measure how close to diagonal it is by taking the Frobenius norm of the resulting matrix without the diagonal. Something like $ \| off( X^{-1} A X ) \|_F / \| A \|_F $ so that it is a relative measure. If the basis consists of the eigenvalues, then the measure yields $ 0 $. So, the smaller, the better. $\endgroup$ May 29, 2018 at 6:23