If $\lambda$ is a characteristic root of a non-singular matrix $A$, then prove that $\det(A)/\lambda$ is a characteristic root of $\mathrm{Adj}(A)$.

Question

Let $A$ be a $n\times n$ matrix. Then the characteristic equation is $\mathrm{det}(A-\lambda I)=0$, with possibly characteristic roots of $A$ $\lambda_1,\dots,\lambda_n$. Let $A$ be regular, i.e. $\det(A)\neq0$. Now $\mathrm{Adj}(A)=\det(A)\mathrm{Inv}(A)$. I don't know how to find its characteristic roots.

Answer 1

Let $A$ be invertible. As you rightly remarked, then $\mathrm{adj}(A)=\mathrm{det}(A)A^{-1}$. We can even prove the following stronger proposition:

Proposition: For $A$ invertible, $\lambda\neq 0$ is an eigenvalue of $A$ iff $\mathrm{det}(A)\lambda^{-1}$ is an eigenvalue of $\mathrm{adj}(A)$:

Proof: $$ Av=\lambda v\Leftrightarrow \mathrm{adj}(A)(Av)=\mathrm{adj}(A)(\lambda v)\Leftrightarrow \mathrm{det}(A)A^{-1}(Av)=\mathrm{det}(A)A^{-1}(\lambda v)\Leftrightarrow \mathrm{det}(A)v=\mathrm{det}(A)\lambda A^{-1}(v)\Leftrightarrow\mathrm{det}(A)\lambda^{-1}v=\mathrm{det}(A)A^{-1}(v)\Leftrightarrow\mathrm{det}(A)\lambda^{-1}v=\mathrm{adj}(A)v $$ $\Box$

You may rephrase this result in terms of roots of the characteristic polynomial, as you naturally have for $\lambda\neq 0$ that $\mathrm{det}(A-\lambda E_n)=0$ iff $\lambda$ is an eigenvalue of $A$ iff $\mathrm{det}(A)\lambda^{-1}$ is an eigenvalue of $\mathrm{adj}(A)$ iff $\mathrm{det}(\mathrm{adj}(A)-\mathrm{det}(A)\lambda^{-1}E_n)=0$.

Obviously, the condition $\lambda\neq 0$ is w.l.o.g. as $0$ is a eigenvalue of $A$ iff $\det(A-0E_n)=\det(A)=0$ iff $A$ is singular.

Answer 2

Let $B:=\det(A) A^{-1}$. For a scalar $\mu \ne 0 $ and a vector $x \ne 0$ we have

$Bx= \mu x \iff \det(A)A^{-1}x= \mu x \iff \det(A)x= \mu Ax \iff Ax= \frac{\det(A)}{\mu}x.$

Answer 3

$$Au= \lambda u \Rightarrow \mbox{Adj}(A)Au = \mbox{Adj}(A)\lambda u \Rightarrow \det(A) u = \mbox{Adj}(A)\lambda u \\ \Rightarrow \frac{\det(A)}{\lambda} u = \mbox{Adj}(A) u$$

Alternately note that $$\det(xI-\mbox{adj}(A))= \det(x\frac{1}{\det(A)}A \mbox{adj}(A)-\mbox{adj}(A))\\= \det(x\frac{1}{\det(A)}A -I) \det(\mbox{adj}(A))= (\frac{x}{\det(A)})^n \det(A - \frac{\det(A)}{x}I) \det(\mbox{adj}(A))\\= (\frac{1}{y})^n \det(A - yI) \det(\mbox{adj}(A))$$ where $y=\frac{\det(A)}{x}$.