How to find when an eigenbasis

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This is a general question because I don't seem to be getting a straight answer from my textbook. I know how to find eigenvalues/vectors, but finding IF an eigenbasis exists is what is throwing me off.

Can someone give me an example of, say, a 3x3 matrix and go through the procedure of finding if the basis exists and then how to find it?

I know this is probably asking a lot, but I am confused!

Any help would be much appreciated.
 
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Nevermind, I feel dumb for posting the question! I found some examples :-)
 
Consider the characteristic equation det(A-λI)=0.

It factors in ℂ as (λ-a)(λ-b)(λ-c)=0.

If a,b,c are distinct and real, then you have an eigenbasis.

If a,b,c are distinct, but b and c are complex conjugates, then a has an eigenvector, while b and c have complex [STRIKE]eignevalues[/STRIKE] eigenvectors, which are commonly adjusted so that A is the rotation of some plane orthogonal to the eigenvector of a.

If a is disticinct from b=c, then a has an eigenvector, while b=c has either one or two eigenvectors. If it has one, than there is something called a generalized eigenvector, which you might read about that, and the jordan form, which is a generalization of diagonalizing a matrix.

If a=b=c, then there are either 1, 2 or 3 eigenvectors, and respectively either 2, 1, or 0 generalized eigenvectors.

An example of a matrix with 1 eigenvalue and 3 eigenvectors is the identity matrix.

I=(1,0,0)(0,1,0)(0,0,1)

An example of a matrix with 1 eigenvalue and only 1 eigenvector is

A=(1,1,0)(0,1,1)(0,0,1)

Proof is left as an exercise for the reader.
 
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Thank you for your detailed reply. This is helping a lot.
 
Just making sure of this, but... we need to make sure that the multiplicity of the eigenvalue (if only one) equals n of I_n.

I'm doing some late-night studying, catching up so hopefully I can pass my exam on Wednesday. I hope I can learn to like linear algebra after this class is over. I'm aware of its many applications in engineering, statistics, etc.
 
In the charactersitic equation, each eigenvalue will have an algebraic mulitplicity (i.e.(λ-a)k). An eignevalues algebraic multiplicity is at least as large as its geometirc multiplicity (the number of eigen vectors it has (or the dimension of its eigenspace) (alg≥geom)). If the algebraic is not equal to the geometric (alg>geom), then some call this degeneracy. You can "fix" this degeneracy by finding generalized eigenvectors, which are not strictly speaking eigenvectors, but they are nice enough, and we will find a useful basis after all.

So I think you mean the geometric multiplicity of each eigenvalue should equal their corresponding algebraic multiplicities, to get a full set of eigenvectors. Something like that, I haven't thought carefully about the complex, maybe they always get a nice rotation interpretation (each conjugate pair).

Any yes, enjoy it, one of my favorite ways to appreciate linear algebra is that it is the most easily analyzed case of the obviously practical subject of maps from one collection of data to another collection of data, or a transform from one signal, through an operator to another signal, like in a circuit. Blah blah blah, multivariable math is cool!
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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