Math, asked by buvibuvanesh1952003, 4 months ago

how to find eign vectors if three eign values are same​

Answers

Answered by TheBrainlyKing1
0

Step-by-step explanation:

If, by similar, you mean ‘the same’, then you have a degenerate eigenspace.

It might be useful to take a geometrical analogy. The eigenvectors will span the vector space, meaning you can express any vector in the space as a combination of the eigenvectors. Your Vector Vi will be expressed as {v1,v2,v3,v4…} where v1, v2, .. are normalized (length one) eigenvectors with corresponding eigenvalues (ei,e2,e3…) When you operate on the vector with a corresponding matrix, instead of having to do N^2 operations, you just end up with {e1vi, e2v2,e3v3,… } which is N multiplications.

What does it mean if two vectors have the same eigenvalue? It means that any linear combination of those two vectors is also an eigenvector with the same eigenvalue. Pick any one you want. Pick one that makes it easy to work with in the system you use. The good news is, that since there are two of them, you can pick two that have the same eigenvalue, AND are orthogonal to each other (just subtract the projection of one onto the other from it, and the residue, by definition, will be orthogonal.

If, say, you’re in 3D space, and all three eigenvalues are the same, what this means is that you have three eigenvectors in a 3D space, so pick any three orthogonal vectors in the space that you want and they’ll be orthogonal eigenvectors under the operation.

Or say you had an operation that turned any vector (A,B,C) into (3A,2B,3C) . You have one eigenvector (0,1,0) with eigenvalue 2. Then any vector in (A,0,C) is an eigenvector with eigenvalue 3. You could pick (1,0,0) and (0,0,1). Or (1/sqrt(2))(1,0,1) and (1/sqrt(2))(1,0,-1) . Or any other pair of orthogonal vectors in that 2-space.

The eigen-vector finding process can go a lot of ways. If you have similar values, it may be simplest to go find the vectors, either by construction, or with methods like Gram-Schmidt, or numerically with an iterative application of the method of dominant eigenvalues.

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