-1
$\begingroup$

I recently came across a problem:

"Normalise the three eigenvectors of the matrix $B$." The matrix $B$ was given by: $$ B = \begin{bmatrix} 2 & 0 & i\\ 0 & 1 & 0\\ i & 3 & 2 \end{bmatrix}. $$

I found the three eigenvalues and corresponding eigenvectors, and found two of them to be orthogonal.

My question is, does "Normalise" mean to convert these eigenvectors to unit vectors, or to form an orthonormal basis from them (for example, use the Gram-Schmidt procedure)?

$\endgroup$
  • 1
    $\begingroup$ Are you sure you're not missing a minus sign somewhere? In QM, matrix representations of observables are always hermitian. Of course, this does not have to be the case here. $\endgroup$ – noah May 11 '17 at 12:39
  • $\begingroup$ Problem is from which reference? $\endgroup$ – Qmechanic May 12 '17 at 6:30
  • $\begingroup$ The Gram-Schmidt procedure, when applied to the $n$ eigenvectors of an operator, does not generally result in another set of $n$ eigenvectors. (Unless the eigenvectors are already orthogonal, which will be the case for a Hermitian matrix.) $\endgroup$ – Michael Seifert May 12 '17 at 15:01
0
$\begingroup$

If we calculate eigenvectors $\vec{s}_j$ of a matrix $\mathbf{B}$, they are solutions to the equation $$ \mathbf{B}\vec{s}_j = \lambda_j \vec{s}_j$$ From the above equation you can clearly see, that eigenvectors are only determined up to a factor, since $\vec{q}_j = a\vec{s}_j$, with an arbitrary scalar $a \neq 0$, also solves the above statement which you can see by simply multiplying the whole equation by $a$. We can use this degree of freedom to ensure that $||\vec{s}_j|| = 1$, which is usually called normalization.

However, linear combinations of eigenvectors of different eigenvalues - which is what a Gram-Schmidt othogonalization would do to the vectors in your example - are not eigenvectors of the matrix. $$ (a\vec{s}_j + b\vec{s}_k) \mathbf{B} = a \lambda_j \vec{s}_j + b \lambda_k \vec{s}_k \neq \lambda (a\vec{s}_j + b\vec{s}_k)$$ So orthonormalizing them would break them being eigenvectors. Notice that this is not the case if the eigenvalue is degenerate, i.e. both eigenvectors have the same eigenvalue. This is easy to see in the above equation when setting $\lambda_j = \lambda_k$.

As someone stated above, if the matrix were hermitian, you can always choose the basis vectors to be orthonormal. The term choose comes from possible degeneracies, in which subspace you can in fact orthonormalize - so make them orthogonal and scale them to have norm 1 - the vectors against each other. If there are no degenerecies, the eigenvectors are automatically orthogonal.

$\endgroup$
1
$\begingroup$

In the text books I have seen to "normalise" a vector would mean to ensure it have norm one, i.e. is a unit vector.

Also, if you used the Gram-Schmidt procedure you wouldn't necessarily end up with eigenvectors of the matrix.

$\endgroup$
  • 1
    $\begingroup$ I would add that $B$ is not hermitian. Only a hermitian matrix guarantees that you can find an orthogonal basis of eigenvectors. $\endgroup$ – Noiralef May 11 '17 at 10:37

Your Answer

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