3
$\begingroup$

Why is it that the characteristic polynomial for a matrix $A$

$$\phi(\lambda) = \det(\lambda I - A)$$

when finding the roots gives the eigenvalues of $A$?

$\endgroup$
  • $\begingroup$ because $A x = 0$ has non trivial solutions iff $det(A) = 0$ $\endgroup$ – vnd Jul 27 at 1:01
  • $\begingroup$ but why would you want $Ax = 0$? Shouldn't $Av = \lambda v$? $\endgroup$ – Vahan Jul 27 at 1:09
  • $\begingroup$ Yes, but apply the fact @vnd mentioned to $A-\lambda I$. $\endgroup$ – Chris Custer Jul 27 at 3:35
6
$\begingroup$

By definition, an eigenvalue of a matrix $A$ is a number $\lambda$ such that $Ax=\lambda x$ where $x$ is a non zero vector, called the eigenvector corresponding to $\lambda$. This means $(A-\lambda I)x=0$ for this $x$. That is, $\lambda$ is an eigenvalue of $A$ iff $A-\lambda I$ is not invertible. That is, $\lambda$ is an eigenvalue of $A$ iff $\det (A-\lambda I)=0$. Here $\det (A-\lambda I)$ is a polynomial in $\lambda$ and any $\lambda$'s satisfying this polynomial are eigenvalues!

$\endgroup$
  • $\begingroup$ Why is it that $A - \lambda I$ must not be invertible? $\endgroup$ – Vahan Jul 27 at 3:32
  • $\begingroup$ Since, $(A-\lambda I)x=0$ means there is a non zero vector $x$ in the null space of $A-\lambda I$, concluding $A-\lambda I$ is not one-one and hence not invertible $\endgroup$ – Chinnapparaj R Jul 27 at 3:38
3
$\begingroup$

$\lambda$ is an eigenvalue of $A$ iff $A-\lambda I$ has a non-trivial null space, which is true iff $\det(A-\lambda I)=0$, which is equivalent to $p(\lambda)=0$, where $p$ is the characteristic polynomial of $A$.

$\endgroup$
2
$\begingroup$

We need a non zero vector $V$ to satisfy $$AV=\lambda V$$

Thus the homogeneous system $$(A-\lambda I)V=0$$ must have a nontrivial solution.

Thus we have to have $$det(A-\lambda I)=0$$

$\endgroup$
0
$\begingroup$

We want to find all $\lambda $ satisfuing $Ax= \lambda x \rightarrow (\lambda x -Ax)=0 \rightarrow(\lambda I -A)x=0 ; x \neq 0$ Therefore $ \lambda I -A$ is singular, meaning has determinant $0$ . Expanding on the determinant, we find all $\lambda$ that satisfy this condition. Was this clear/helpful?

$\endgroup$
  • $\begingroup$ This is not clear, because $x$ is being used without being introduced, in particular without any quantifier. $\endgroup$ – Carsten S Jul 27 at 12:53
  • 2
    $\begingroup$ @CarstenS $x$ is "quantified" by "$x \ne 0$" which applies to the whole chain of inferences. $\endgroup$ – alephzero Jul 27 at 15:46
  • $\begingroup$ @alephzero, does that mean "for all $x\ne0$", "for some $x\ne0$", "for my favorite $x\ne0$"? Also note that $\lambda$ has a quantifier, and the order matters. This kind of sloppiness is harmful when communicating with a beginner. Also note that "therefore" usually denotes implication in only one direction, which is less than what is wanted here. $\endgroup$ – Carsten S Jul 31 at 9:43
  • $\begingroup$ @CarstenS: I dont see the need to quantify for $x \neq 0$ We want to find a solution to $Ax = \lambda x =0 $ for $ x \neq 0$. Just what else needs to be quantified? How is that sloppy? Where is the ambiguity? $\endgroup$ – MSIS Aug 2 at 0:02

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.