1
$\begingroup$

For a symmetric matrix $A$, I am aware that eigenvectors $v_1, \dots, v_n$ with the same eigenvalue $\lambda$ are linearly independent but not orthogonal.

The spectral theorem states that any $p \times p$ symmetric matrix has $p$ orthonormal eigenvectors. I do not understand how both these statements are correct when an eigenvalue can correspond to multiple eigenvectors and thus these eigenvectors can only be linearly independent with one another, and not orthogonal?

$\endgroup$
2
  • 3
    $\begingroup$ The spectral theorem states that any symmetric matrix has a basis of orthonormal eigenvectors, but not that every possible eigenbasis will be orthogonal. Note, for instance, that the size $n$ identity matrix is symmetric but every basis for $\Bbb R^n$ is an eigenbasis. $\endgroup$ – Ben Grossmann Oct 25 '18 at 23:53
  • $\begingroup$ "eigenvectors $v_1,\ldots,v_n$ with the same eigenvalue $\lambda$ are linearly independent..." That's certainly not true in general. For example, if $v$ is an eigenvector for $\lambda$, then so is $\alpha v$ for any nonzero scalar $\alpha$, but $v$ and $\alpha v$ are obviously not linearly independent. Similarly, if $v_1$ and $v_2$ are eigenvectors for $\lambda$, then so is any nonzero linear combination of $v_1$ and $v_2$ $\endgroup$ – Bungo Oct 26 '18 at 1:11
1
$\begingroup$

You can get an orthonormal basis from the $v_1,\dots,v_n$ using Gram-Schmidt, say. Then, if $v_1',\dots, v_n'$ is such a basis, it is easy to see that each $v_i'$ is still an eigenvector for $\lambda $. This basis satisfies the requirements of the spectral theorem.

$\endgroup$

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.