MATLAB: Why do I get different EigenVectors than the instructor got
eigenvectorslinear algebra
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As has been pointed out several times, this matrix is not diagonalizable. eig can go as far as to find matrices V and D such that A*V-V*D is effectively zero, to within reasonable tolerances. But there is no matrix V such that V is an orthogonal matrix, where D=V'*A*V, and D is diagonal.
A*V - V*Dans = -4.5475e-12 -2.7285e-12 9.0949e-13 2.7285e-12 1.263e-12 6.8212e-13 9.0949e-12 5.457e-12 2.2737e-12 -1.7053e-13 -1.8066e-12 -1.08e-12 -5.457e-12 3.4106e-13 9.0949e-13 2.2737e-13 -7.6739e-13 -3.4106e-13 0 -5.457e-12 -3.1832e-12 -1.819e-12 4.9738e-14 -4.8317e-13 4.5475e-13 -9.0949e-13 0 -1.4779e-12 1.4388e-13 1.3642e-12 0 -1.3642e-12 -4.5475e-13 2.2737e-12 -2.5757e-13 -1.1369e-12These numbers are all as close to zero as one can expect. But the eigenvector matrix so produced is not orthogonal, not a complete set of eigenvectors. In the sense that an eigenvalue/vector pair satisfies A*v = lambda*v, we can check that for a few eigenvalues just to convince you of that fact...
[A*V(:,1),D(1,1)*V(:,1)]ans = 4739.4 4739.4 -10608 -10608 10904 10904 -7536.4 -7536.4 3265.1 3265.1 -717.44 -717.44[A*V(:,2),D(2,2)*V(:,2)]ans = 4756.5 4756.5 -6853.2 -6853.2 -997.38 -997.38 7704.3 7704.3 -6673.1 -6673.1 2125.5 2125.5This is what it means to say that A*V == V*D.
Anyway, it appears that the determinant test is what may have been bothering you. So if we look at the computed eigenvalues...
E = eig(A)E = 17941 13379 8573.2 4213.6 31.232 1220.7See that we can shift the eigenvalues.
eig(A - E(5)*eye(6))ans = 17910 13347 8542 4182.4 1.5323e-12 1189.5which leaves one of those eigenvalues essentially zero. HOWEVER, the determinant, which could be computed using the product of the eigenvalues, is not zero. Of course, this is to be expected, since even though that one element is tiny, it is NOT sufficiently small when you take the product of those numbers.
prod(ans)ans = 1.5566e+07Again, this is why one should NEVER use det for ANY numerical test. See that rank recognizes the matrix as being singular.
rank(A - E(5)*eye(6))ans = 5In fact, det in MATLAB is computed using an LU factorization, which gives aslightly different value for the determinant.
[L,U,P] = lu(A - E(5)*eye(6))L = 1 0 0 0 0 0 -0.97155 1 0 0 0 0 0 -0.97404 1 0 0 0 0 0 -0.98044 1 0 0 0 0 0 -0.98517 1 0 0 0 0 0 -0.99034 1U = 5105.8 -5720.2 0 0 0 0 0 5065 -4933.5 0 0 0 0 0 4950.3 -4853.5 0 0 0 0 0 4295.4 -4231.6 0 0 0 0 0 3263.5 -3232 0 0 0 0 0 9.0949e-13P = 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1prod(diag(U))ans = 1.6322e+06Compare that result to what det gives:
det(A - E(5)*eye(6))ans = 1.6322e+06Again, DON'T USE DET! The factor of roughly 10 difference between that and what we get from the product of the eigenvalues just emphasizes the reason why you should never use det for these things. That number IS zero, as close as we can come to zero in terms of floating point arithmetic.
For the perfectionists among you, lets try it in symbolic form.
[V,D] = eig(sym(A));det(A - D(5,5)*eye(6))ans =-0.000000000000000062420635190136154950824971272636Still not zero, but closer to that value.
eig(sym(A) - D(5,5)*eye(6))ans = 4562.3474811932686073791772754019 -1.2794464375961745433981454406958e-36 -4805.4337708041558234015199714421 -9165.0606437451593865984338324246 -12157.955591367554507466929334176 -13347.448098888493727903150725422Hi,
The eigenvalues and right-side eigenvectors are correct, they are just returned in a different order than you assume:
>> syms a b d>> A = [a b; 0 d]A =[ a, b][ 0, d]>> [V, D, P] = eig(A)V =[ 1, -b/(a - d)][ 0, 1]D =[ a, 0][ 0, d]P = 1 2The column vectors of V are r1 and r2 from the paper, just with a different scaling for r2.
The problem with the third output is that for regular MATLAB matrices, this would be the left-side eigenvectors of the matrix. However, for symbolic matrices, it's a vector of indices to the linearly independent eigenvectors (see here )
To get the left eigenvalues of A, simply compute the right eigenvectors of A transposed:
>> [W, D] = eig(A.')W =[ (a - d)/b, 0][ 1, 1]D =[ a, 0][ 0, d]
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