Can an eigenvector belong to several eigenvalues

 
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HDMIii
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With us since: 03/21/2017
Messages: 165
Topic start: 2019-04-11

Hello, I have found the following phrases. I have a few questions or would like to ask you if my considerations are correct. I hope you can understand my thoughts. In the following I write y (i.e. y = lambda) for the eigenvalues ​​lambda. 1. In the simplest case I would interpret the theorem as follows: If, for example, a 3x3 (general nxn) matrix has 3 (n) pairwise different eigenvalues ​​y1, y2 and y3, the eigenvectors (which belong to the eigenvalues: y1-> v1 , y2-> v2, y3-> v3) linearly independent. I.e. I have 3 (n) linearly independent eigenvectors. But what if not all eigenvalues ​​are pairwise different or an eigenvalue has 2 linearly independent eigenvectors? My thoughts on this below: What happens if, for example, a 3x3 (nxn) matrix has only 2 (n-1) eigenvalues ​​y1 and y2, where both are different in pairs. Then one can say that the eigenvector v1 (of y1) and v2 (of y2) are linearly independent. But you couldn't say that the matrix only has 2 (n-1) linearly independent eigenvectors, right? What if the characteristic polynomial of a 3x3 (nxn) matrix has a factor with the algebraic multiplicity of e.g. 2. Then two eigenvalues ​​e.g. y1 and y2 have the same value. That means y1 and y2 would no longer be different in pairs. We would then no longer have 3 (n) different eigenvalues, but only 2 (n-1) different eigenvalues. Then one can say that the eigenvector v1 (of y1 or y2) and v3 (of y3) are linearly independent. But you couldn't say that the matrix only has 2 (n-1) linearly independent eigenvectors, right? What if an eigenvalue has 2 linearly independent eigenvectors (i.e. they have the geometric multiplicity 2). In this case, are these 2 eigenvectors (or is the plane spanned by these 2 eigenvectors then also linearly independent of all other eigenvectors of the other paired eigenvalues? That would mean, however, that if I have 2 pairwise different eigenvalues ​​y1 (geometric multiplicity 2) and y2, then I would have 2 pairwise different eigenvalues ​​(the index m in the formula would run up to y2) but 3 linearly independent eigenvectors (the index m would run to 3), which according to the same indices of y and v (both run to m) cannot be. 2. It is logical that a vector space K ^ n can only be spanned with n linearly independent vectors (e.g. this is not possible with n-1 linearly independent vectors). Now we know from 1. that a matrix with n pairwise different eigenvalues ​​certainly has n linearly independent eigenvectors and these eigenvectors therefore certainly form a basis of K ^ n. But you CANNOT say, as soon as there are only n-1 pairwise different eigenvalues ​​(if one has e.g. the algebraic multiplicity 2 or the characteristic polynomial has a 3x3 matrix e.g. only 2 solutions) there are no longer n linearly independent eigenvectors and therefore can the eigenvectors no longer form the basis of K ^ n (but there can still be n linearly independent eigenvectors, since an eigenvalue can also have several eigenvectors). 3. i) That would mean as soon as A only has n-1 linearly independent eigenvectors, the eigenvectors no longer form the basis of K ^ n (see explanation under 2.). Then A is no longer diagonalizable. ii) is clear 4. It follows from 3i) and 2. The statement is: Every matrix A is diagonalizable which has n pairwise different eigenvalues. However, one cannot say that the matrix cannot be diagonalized if it has fewer than n pairwise different eigenvalues ​​(since fewer than n pairwise different eigenvalues ​​does not mean that there are necessarily fewer than n linearly independent eigenvectors, since one eigenvalue also has several eigenvectors may own). https://matheplanet.com/matheplanet/nuke/html/uploads/b/47643_jjjjj.png


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Creasy
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With us since: 02/22/2019
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Place of residence: Bonn
Post No.1, registered 2019-04-12

Hey, in general, you often speak of "the" eigenvector of an eigenvalue. One would rather say "one" eigenvector, because (mostly) there are very many eigenvectors (all multiples of an eigenvector, for example). Now to 1): Yes, if you have a 3x3 matrix with 3 different eigenvalues ​​and an eigenvector for each eigenvalue, then these eigenvectors are l.u. (that's the statement) \ quoteon What happens if, for example, a 3x3 (nxn) matrix has only 2 (n-1) eigenvalues ​​y1 and y2, where both are different in pairs. Then one can say that the eigenvector v1 (of y1) and v2 (of y2) are linearly independent. But you couldn't say that the matrix only has 2 (n-1) linearly independent eigenvectors, right? \ quoteoff Yes, an eigenvector for $ \ lambda_1 $ and an eigenvector for $ \ lambda_2 $ are then 1.u .. And yes, one cannot make a statement about whether there are three linearly independent eigenvectors. The next section is exactly the same? \ quoteon What if an eigenvalue has 2 linearly independent eigenvectors (i.e. they have the geometric multiplicity 2). In this case, are these 2 eigenvectors (or is the plane spanned by these 2 eigenvectors then also linearly independent of all other eigenvectors of the other pairwise different eigenvalues? \ quoteoff Yes, if one finds two linearly independent eigenvectors $ v_1, v_2 $ for an eigenvalue and also finds eigenvectors $ v_3, \ ldots, v_n $ for further different eigenvalues, then $ v_1, v_2, v_3, \ ldots, v_n $ are linear independently. To show this, you can look at the evidence of the first statement. 2. Do you understand correctly. 3. Correct: If one finds only $ n-1 $ -many linearly independent eigenvectors for an nxn matrix $ A $, then $ A $ is not diagonalizable. Here you write: \ quoteon the eigenvectors no longer form a basis of K ^ n \ quoteoff and mean: one cannot find a basis of $ K ^ n $ from eigenvectors. All eigenvectors together (almost) never form a basis of $ K ^ n $, since most of them are multiples of each other (i.e. linearly dependent). The question is whether you can choose which ones to form a basis. 4. Do you understand correctly. Here again as an example: \ quoteon because an eigenvalue can also have several eigenvectors \ quoteoff If you have found an eigenvector, then all multiples of it are also eigenvectors (with the same eigenvalue). What you mean is something like: The dimension of the eigenspace does not necessarily have to be one. So pay attention to the language a little, otherwise I hope I could help. Best regards, Creasy


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HDMIii
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With us since: 03/21/2017
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Post No.2, from the topic starter, entered 2019-04-12

Ok, thank you very much for your effort :)


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