Unable to find out left eigen vectors of symbolic matrix

Hello,

I have 3*3 matrix having symbolic elements. I tried to find out the eigen value, left eigen vector and right eigen vector using below command

[right_eig_vectors, eig_values] = eig(A);
[left_eig_vectors, ~] = eig(A.');

My understanding is that the left eigenvectors solve the following equation: W'*A-D*W'=0. But when I put that into solve ,it partially solve this equation of W'*A-D*W'=0. Can anyone help me?

[Christine Tobler]

Possibly the eigenvalues returned for eig(A) and for eig(A.') are in sorted in a different order? When I tried for an example, the order of the eigenvalues matched, but this might not be the case for your input A.

[Walter Roberson]

Looks solved to me?

Reminder: Using .' suggests that you are using real-valued A, but you did not tell us that. My understanding is that if A is potentially complex valued, you shoul be taking eig(A')

syms A [3 3] 
[right_eig_vectors, eig_values] = eig(A)

right_eig_vectors = 

eig_values = 

[left_eig_vectors, ~] = eig(A.')

left_eig_vectors = 

[Abhinav Kumar]

Hi,

I have attached the matlab code file

syms z1 z2 z3 z4 z5 z6 z7 ;

A=[z1 z2 z3; z4 z5 z6; 1 2 z7]

M=subs(A,[z1 z2 z3 z4 z5 z6 z7],[-0.1 0.1 -2 0.01 0.07 -0.5 0.2])

[V,D] = eig(M)

[W, ~] = eig(M.')

W'*M - D*W'

The W'*M-D*W' should be equal to or almost zero but that not the case in my coding.

The W'*M-D*W' outputs are like this

ans =

1.0e-14 *

-0.1388 0.5329 0

ans =

1.9774 + 0.1985i 4.0123 - 0.0596i 0.0000 + 3.4528i .(this should be equal to zero)

ans =

1.9774 - 0.1985i 4.0123 + 0.0596i 0.0000 - 3.4528i. (This should be equal to zero)

Is there any different command which I can use for correct left eigen vectors or need any modification

in this command. Please help

Answer 1

[Christine Tobler]

The left eigenvectors are still expressed as right eigenvectors of M', meaning they satisfy a slightly different equation:

syms z1 z2 z3 z4 z5 z6 z7  ;
A=[z1 z2 z3; z4 z5 z6; 1 2 z7];
M=subs(A,[z1 z2 z3 z4 z5 z6 z7],[-0.1 0.1 -2 0.01 0.07 -0.5 0.2]);
[V, D] = eig(M);
[W, D2] = eig(M');

assert(isequal(D, D2))
double(norm(M*V - V*D))

ans = 0

double(norm(M'*W - W*D))

ans = 0

% Apply conjugate transpose to the second expression:
double(norm(W'*M - conj(D)*W'))

ans = 0

% Apply non-conjugate transpose instead:
double(norm(W.'*M - D*W.'))

ans = 0

% You can also apply CONJ to W, in which case the formula becomes:
W = conj(W);
double(norm(W'*M - D*W'))

ans = 0

% This last one matches what happens in EIG for floating-point numbers,
% where it's possible to compute left and right eigenvalues in one go.
Md = double(M);
[Vd, Dd, Wd] = eig(Md);
norm(Md*Vd - Vd*Dd)

ans = 1.9481e-15

norm(Wd'*Md - Dd*Wd')

ans = 9.5794e-16

So in short, taking the conjugate of the matrix W should resolve your issues.

[Abhinav Kumar]

Thank you very much Sir for your answer