In which cases does the alg multiplicity differ from geometric and why? In the cases where the quadratic nxn Matrix does transform on a lower dimensional subspace (det=0) it‘s clear, the Eigenvectors can not span the whole n dim space but in this example at the beginning even though we have full rank, the eigenvectors don‘t span the whole R3. Why not?
The geometric multiplicity is the dimension of the nullspace of A - lambda*I not the dimension of the nullspace of A. Matrix A can be full rank (say n) and yet A - lambda*I may have rank n -1 or n -2 etc etc. The eigenvectors (v) corresponding to eigenvalue lambda satisfy the nullspace equation (A - lambda*I ) * v = 0. If rank A - lambda*I = n-1 then the nullspace has only a single eigenvector. If rank A - lambda*I = n -2 then the nullspace has 2 eigenvectors for that eigenvalue. This follows from the rank nullity theorem whereby dimension nullspace + rank = n. The matrix A - lambda*I is ALWAYS singular.
I should add that the rank of A - lambda*I is not necessarily less than the rank of A. In fact they could be identical. E.g., take any singular 2 X 2 matrix with 2 distinct eigenvalues. In that case rank A = 1 and rank A - lambda*I = 1 also. Each eigenspace will have dimension = 1.
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In which cases does the alg multiplicity differ from geometric and why? In the cases where the quadratic nxn Matrix does transform on a lower dimensional subspace (det=0) it‘s clear, the Eigenvectors can not span the whole n dim space but in this example at the beginning even though we have full rank, the eigenvectors don‘t span the whole R3. Why not?
The geometric multiplicity is the dimension of the nullspace of A - lambda*I not the dimension of the nullspace of A. Matrix A can be full rank (say n) and yet A - lambda*I may have rank n -1 or n -2 etc etc. The eigenvectors (v) corresponding to eigenvalue lambda satisfy the nullspace equation (A - lambda*I ) * v = 0. If rank A - lambda*I = n-1 then the nullspace has only a single eigenvector. If rank A - lambda*I = n -2 then the nullspace has 2 eigenvectors for that eigenvalue. This follows from the rank nullity theorem whereby dimension nullspace + rank = n. The matrix A - lambda*I is ALWAYS singular.
I should add that the rank of A - lambda*I is not necessarily less than the rank of A. In fact they could be identical. E.g., take any singular 2 X 2 matrix with 2 distinct eigenvalues. In that case rank A = 1 and rank A - lambda*I = 1 also. Each eigenspace will have dimension = 1.