MATLAB: Eig command returning wrong eigenvectors

calculationeigeigenvalueeigenvectorwrong value

Hi!

I am very new to using Matlab, so this question might seem silly.

I have a matrix:

    K =
      -7            -18            -18              9             27       
       0              3              0              0              0       
      -8            -12             -5              7             17       
     -10            -16            -20             10             25       
      -6            -10             -6              6             17

My goal is to calculate the eigenvectors and eigenvalues, which I attempt by using the command

    [M,N] = eig(A)

This command gives my two new matrices, one with the values and the other with the vectors. I get the right results for the values: 2, 3 and 5 (3 and 5 twice) but the vectors aren't correct:

M =
     273/1520         *            647/2153      -509/586          *       
       0              0              0              0            310/409   
    -273/1520       888/2737      2811/5162       266/1651       283/1695  
     273/304      -2220/2737      -847/1906      -452/1003      -293/7624  
    -273/760       1332/2737       813/1261      -446/3473       873/1387

The vectors should instead be something along the lines of (-1/2,0,1/2,-5/2,1).

What am I doing wrong? (I am assuming it's me and not the program)

Thanks in advance!

Best Answer

Doing wrong? First, it looks like you have format set to rat.

[V,D] = eig(K);
V
V =
         0.179605302026751     -5.06266247219048e-14         0.300510985421823        -0.868600674748355      1.49364371455249e-15
                         0                         0                         0                         0         0.757946284651594
        -0.179605302026794         0.324442842261484         0.544556381601175         0.161114353924498         0.166961764835024
         0.898026510133868        -0.811107105653849        -0.444386053127199         -0.45064791217395       -0.0384312697617635
        -0.359210604053568         0.486664263392254         0.644726710075106        -0.128419204324953         0.629415789578334
diag(D)
ans =
          2.00000000000014
          2.99999999999972
          5.00000000000016
                         5
                         3

So eigenvalues of 2,3,5, the latter are of essentially multiplicity 2 each, if we ignore the trash in those least significant bits.

How about the eigenvectors?

You need to recognize that an eigenvector is unique only to within a scale factor. An eigenvector (V) basically has the property

K*V = lambda*V

for corresponding eigenvalue lambda.

But if that holds, then surely

K*(s*V) = lambda*(s*V)

for any scalar value s. So what happens if you scale those eigenvectors?

V(:,1)/V(1,1)
ans =
                       1
                       0
       -1.00000000000024
        5.00000000000064
       -2.00000000000037

How interesting.

So eig is NOT returning the wrong eigenvectors. It merely normalizes the eigenvectors so they have unit norm. Your expectations were wrong.

norm(V(:,1))
ans =
                       1

But that is just scaling the vectors by a constant. And we just discussed that that constant is completely arbitrary. At the same time, having unit normalized eigenvectors is very useful in mathematics.

Related Solutions

MATLAB: Matlab code problem (calculate eigenvalues and eigenvectors)

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*D
ans =
  -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-12

These 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       2125.5

This 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...

See that we can shift the eigenvalues.

eig(A - E(5)*eye(6))
ans =
        17910
        13347
         8542
       4182.4
   1.5323e-12
       1189.5

which 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+07

Again, 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 =
     5

In 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            1
U =
       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-13
P =
     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     1
prod(diag(U))
ans =
   1.6322e+06

Compare that result to what det gives:

det(A - E(5)*eye(6))
ans =
   1.6322e+06

Again, 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 =
-0.000000000000000062420635190136154950824971272636

Still 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
     -13347.448098888493727903150725422

MATLAB: Problem with for loops

Jaime - please don't dynamically create variables. See Stephen's post at https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval.

In your code, you are creating ten arrays of long (should rename this variable to make it more clear what it represents). This is then equivalent to a single 10 x m array (where m is the length of your X) which should be just

 v_a = normrnd(0,1,10,length(X));

Then, in your for loops, you can just reference the appropriate column of v_a instead of trying to figure out which array variable you will need.

Best Answer

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MATLAB: Matlab code problem (calculate eigenvalues and eigenvectors)

MATLAB: Problem with for loops

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