I am assigned to compute eigenvalues and eigenvectors in MATLAB of a 2x2 matrix: $$ A = \left( \begin{matrix} 3 &0\\ 4 &5\\ \end{matrix} \right) $$

I know that the textbook's solution states that eigenvalue 3 corresponds to an eigenvector $(1 \; -2)$, and eig 5 corresponds to $(0 \; 1)$.

This is what I do. Why am I not getting the correct eigenvectors?

% Define the matrix
A = [3 0;4 5];

% Find Eigenvalues
E1 = eig(A);

% Display Eigenvalues
disp('Eigenvalues of the matrix A:')

% Determine Eigenvectors
[V,D] = eig(A);

% V1 corresponds to eigenvalue 1 and V2 corresponds to eigenvalue 2
V1 = V(:,1)
V2 = V(:,2)
  • 1
    $\begingroup$ A tip: If you indent lines by 4 spaces, they'll be marked as a code sample. You can also highlight the code and click the "code" button (with "{}" on it). There's even syntax highlighting! $\endgroup$ – Mauro Vanzetto Sep 18 '17 at 10:07
  • 1
    $\begingroup$ A tip: You can use MathJax to typeset your mathematical formulas. This will make the question much easier to read. $\endgroup$ – Mauro Vanzetto Sep 18 '17 at 10:08

Eigenvectors are only unique to within a scale factor (can be + or - scale factor).

If $x$ satisfies $Ax=\lambda x$, and hence is an eigenvector of $A$ corresponding to eigenvalue $\lambda$, then any multiple of $x$ also satisfies the equation, and hence is also an eigenvector of $A$ corresponding to eigenvalue $\lambda$.

MATLAB normalizes eigenvectors to have 2-norm equal to 1, but even that leaves a choice of sign.

  • $\begingroup$ Of course, if the scale factor is zero, producing a zero vector, then that satisfies the equation, but is not an eigenvector (which must not be the zero vector). But any eigenvector multiplied by a non-zero scale factor is also an eigenvector. $\endgroup$ – Mark L. Stone Sep 18 '17 at 2:40

I doubt you're doing anything wrong. Matlab (and LAPACK, the guts underneath Matlab) will normalize eigenvectors to unit length, so you won't get (1,-2) for an eigenvector, you'll get (1,-2)/sqrt(5) instead.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.