how to find lambda from this formula?
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det (A - lambda * I) = 0
where:
A= a square matrix, ex: [2 5 4;8 6 9;3 4 6]
I= identity matrix
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Brian B
2013-1-3
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e = eig(A) returns a vector of eigenvalues of the matrix A. To get the associated eigenvectors, use [V,D] = eig(A). See the documentation for eig to interpret V and D.
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eri
2013-1-3
Brian B
2013-1-3
There are, in general, multiple solutions (values of lambda) which satisfy the equation det(A-lambda*I)=0. eig(A) returns a vector, each element of which is one solution.
For an n-by-n matrix A, the vector returned by eig(A) will have length n. That is because det(A-lambda*I) is a degree-n polynomial, and hence has n (possibly complex) roots by the fundamental theorem of algebra. Note that some values may be repeated; that indicates the algebraic multiplicity of the root.
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