9
$\begingroup$

"Eigenvectors corresponding to different eigenvalues are linearly independent."

My professor told us this during a lecture, but gave no proof or explanation.

$\endgroup$
0

1 Answer 1

11
$\begingroup$

Suppose that $Av=\lambda v$ and $Aw=\mu w$, $v,w\not=0$.

Assume $v,w$ are linearly dependent, then $v=c\cdot w$ for some scalar $c\not=0$. Then

$$\lambda v=Av=cAw=c \mu w = \mu v$$

That is, $\lambda=\mu$.

$\endgroup$
13
  • 1
    $\begingroup$ You surely mean dependent? $\endgroup$
    – alex
    Mar 2, 2014 at 15:47
  • $\begingroup$ @DanielFischer, alex Thank you! $\endgroup$
    – J.R.
    Mar 2, 2014 at 15:50
  • 2
    $\begingroup$ The theorem is more general then this. If you have $\lambda_1,\ldots,\lambda_m$ as distinct eigenvalues and $v_1,\ldots,v_m$ be corresponding eigenvectors, then $v_1,\ldots,v_m$ are linearly independent. $\endgroup$ Mar 2, 2014 at 15:51
  • 1
    $\begingroup$ @gt6989b I disagree but may be that's just me. $\endgroup$ Mar 2, 2014 at 16:10
  • 5
    $\begingroup$ @caffeinemachine Maybe I don't understand something. If you let $v_1$ be a linear combination of the others, i.e. $$v_1 = \sum_{k=2}^m c_k v_k$$ then $$\sum_{k=2}^m c_k \lambda_k v_k = A \sum_{k=2}^m c_k v_k = A v_1 = \lambda_1 v_1 = \lambda_1 \sum_{k=2}^m c_k v_k = \sum_{k=2}^m \lambda_1 c_k v_k$$ so $$\sum_{k=2}^m \lambda_1 c_k v_k = \sum_{k=2}^m c_k \lambda_k v_k$$ which implies that $\lambda_1 c_k = c_k \lambda_k$ for all $k$, since $v_k$ are linearly independent, and so either $c_k \equiv 0$ or $\lambda_1 = \lambda_k \quad \forall k$. $\endgroup$
    – gt6989b
    Mar 2, 2014 at 17:51

Not the answer you're looking for? Browse other questions tagged .