0
$\begingroup$

I have a doubt that can we find Eigen values and eigen vectors for the linear transformation, I only know to transform linear transformation into matrix form using standard basis or any other basis and finding eigen values and eigen vectors using $|A-\lambda I|$ where $A$ is the matrix representation of the linear transformation.

My question is "Can we find the eigen values and eigen vectors directly from the operator without changing it into matrix form."

$\endgroup$
  • $\begingroup$ That depends on what you know about the operator $\endgroup$ – Vim May 29 '15 at 15:15
  • $\begingroup$ Thanks for the comment.If the transformation is given (say) T(x,y) = x^2 +y^2. can we now find? $\endgroup$ – Sam Christopher May 29 '15 at 15:22
  • $\begingroup$ I don't understand your example. What are $x$ and $y$? I can't think of any interpretation where this is a linear map. $\endgroup$ – Aaron May 29 '15 at 15:24
  • $\begingroup$ @SamChristopher Is the T(x,y) in your example linear? $\endgroup$ – Vim May 29 '15 at 15:32
  • $\begingroup$ And by the way it is an mapping (obviously not linear..) from R^2 to R^1, it is not even a transformation ( Would you care to recall the definition of "linear" and "linear transformation?) $\endgroup$ – Vim May 29 '15 at 15:34
1
$\begingroup$

There is no standard way of presenting a vector space other than by presenting a basis, and so there is no standard way of presenting a linear map between two vector spaces other than a matrix (or something that is equivalent to a matrix, e.g., defining the map on a large enough collection of vectors). So any situation where you are trying to compute eigenvalues and you do NOT have a matrix (or something easy to convert to a matrix) will be slightly contrived.

The easiest example I can come up with is the following: Let $V$ be the collection of polynomials of degree at most $5$, and let $T:V\to V$ be the map sending a polynomial to its derivative. Since the derivative lowers the degree of a polynomial by one (until it is constant, and then equal to zero, at which point the process stabilizes), we have that $T^6$ must be the zero map. This tells us that the minimal polynomial divides $x^6$, and since the characteristic polynomial and the minimal polynomial must share the same roots, the characteristic polynomial must be $x^6$ too. Therefore, $0$ is the only eigenvalue. What are the corresponding eigenvectors? Since differentiation lowers degree by exactly one, only constant polynomials have zero derivative, and the constant polynomials are spanned by the vector $1$.

So here, we had an example of a linear transformation which we could analyze without putting it directly into matrix form because we could find a polynomial it satisfied just by using general properties we knew. So it can be done in some situations. However, doing it in general requires having some way of deducing properties of the transformation, and that usually isn't strictly a matter of linear algebra.

$\endgroup$
2
$\begingroup$

There are two main cases for your question:

1) The description of the operator is specified in such a way that you would be unable to determine its action on any given basis. e.g. The mapping from $\mathbb{R}^2$ to itself that "is a rotation". There are many such rotations, and given the eigenvalues and eigenvectors, you know the transformation. So there the answer is no.

2) The description of the operator is specified in such a way that you would be able to determine its action on any given basis. e.g. The mapping from $\mathbb{R}^2$ to itself that "is a reflection through the $x$ axis". Now the answer to your question is, yes, you can determine its eigenvalues, but no, you cannot uniquely determine its eigenvectors. Given different bases they may be different. However, given a mapping between two bases will map the eignenspaces into each other.

As a final note, a nice application of the concepts in the second description are the heat trace asymptotics for a Riemannian Manifold. Here we look at properties of the eigenvalues of the laplacian operator while not depending on a specific basis. This has the nice property of honing in on geometric invariants of the manifold by removing the particular presentation of the manifold.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.