# Why can the characteristic polynomial be used to find eigenvalues?

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 $$A$$?

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

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!

• Why is it that $A - \lambda I$ must not be invertible? – Vahan Jul 27 at 3:32
• 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 – Chinnapparaj R Jul 27 at 3:38

$$\lambda$$ 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 $$A$$.

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 $$det(A-\lambda I)=0$$

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?

• This is not clear, because $x$ is being used without being introduced, in particular without any quantifier. – Carsten S Jul 27 at 12:53
• @CarstenS $x$ is "quantified" by "$x \ne 0$" which applies to the whole chain of inferences. – alephzero Jul 27 at 15:46
• @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. – Carsten S Jul 31 at 9:43
• @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? – MSIS Aug 2 at 0:02