Geometric Multiplicity: eignvectors (1)

I had a question regarding the relationship between multiplicity of eigenvalue and eigenvectors.

I am more interested in eigenvalue's multiplicity than the value itself. Because if eigenvalue has multiplicity, the number of independent eigenvectors ``could'' decrease. My favorite property of eigen-analysis is that is a transformation to simpler basis. Here, simpler means a matrix became a scalar. I even have a problem to understand a 2x2 matrix, but a scalar has no problem, or there is no simpler thing than a scalar. Ax = λ x means the matrix A equals λ, what a great simplification!

My question is

If λ has multiplicity, are there still independent eigenvectors for the eigenvalue?

My intuition said no. I can compute an eigenvector to a corresponding eigenvalue. But, I think I cannot compute the independent eigenvectors for one eigenvalue.

For instance, assume 2x2 matrix that has λ = 1,1, how many eigenvectors? one?

Recently I found this is related with diagonalization using eigenvector. My intuition was wrong.

For one eigenvalue, that has multiplicity, there can be multiple eigenvectors.

I will show the example of this one eigenvalue and multiple eigenventors in next article.

Comments

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Lx=d: SundayResearch

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