Normalization of the eigenvectors in eig(A) vs. eig(A,eye(length(A)); Orthonormality relationship
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Hello,
Why are the left and right eigenvectors normalized differently in the following two calls of the eig function:
A = magic(3);
B = eye(3);
[V1, L1, W1] = eig(A);
[V2, L2, W2] = eig(A, B);
In addition, from orthogonality of the eigenvectors, we should have:
W'*B*V=eye(3)
However, neither
W1'*eye(3)*V1
nor
W2'*eye(3)*V2
are equal to the identity matrix.
Thank you, Jerome
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Più risposte (1)
David Goodmanson
il 12 Feb 2017
Hello Jerome, It's an interesting question why Mathworks chose not to normalize the column vectors in V2 and W2 when they did so in for V1 and W1, and I'm not attempting to answer that part.
For the other parts of your posting consider just V1 and W1, since introducing normalization factors to the columns of V2 and W2 make them identical to V1 and W1. Keeping things simple, suppose that the eigenvalues are all distinct and nonzero as in this case.
In general neither the set V nor the set W are an orthonormal set of vectors.
>> [V D W] = eig(magic(3))
>> V'*V
1.0000 -0.0000 -0.0000
-0.0000 1.0000 0.3333
-0.0000 0.3333 1.0000
>> W'*W
1.0000 0.0000 0.0000
0.0000 1.0000 -0.3333
0.0000 -0.3333 1.0000
While it is true that each left eigenvector Wi is perpendicular to all but one of the right eigenvectors (call that one Vi), for normalized eigenvectors it is not true in general that Wi ' *Vi = 1. That would mean that W ' *V is the identity matrix, but all that is required is
(W'*V)*D = D*(W'*V)
If D has distinct nonzero eigenvalues then (W ' *V) has to be diagonal but need not be the identity matrix.
>> W'*V
1.0000 -0.0000 -0.0000
-0.0000 0.9428 0.0000
0.0000 -0.0000 0.9428
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