Let $A\in M_{n\times n}(\textbf{F})$. Then a scalar $\lambda$ is an eigenvalue of $A$ if and only if $\det(A - \lambda I_{n}) = 0$.

Question

Proposition

Let $A\in M_{n\times n}(\textbf{F})$. Then a scalar $\lambda$ is an eigenvalue of $A$ if and only if $\det(A - \lambda I_{n}) = 0$.

MY ATTEMPT

We say that $\lambda$ is an eigenvalue of $A$ iff $\lambda$ is an eigenvalue of $L_{A}:\textbf{F}^{n}\to\textbf{F}^{n}$ defined by $L_{A}v = Av$.

On the other hand, $\lambda$ is a eigenvalue of $L_{A}$ iff $(A - \lambda I_{n})v = 0$.

Since $v\neq 0$, it happens iff $\ker(L_{A - \lambda I_{n}})\neq\{0\}$.

This, by its turn, happens iff $L_{A-\lambda I_{n}}$ is not invertible, that is to say, $A - \lambda I_{n}$ is not invertible.

Finally, we arrive at the desired restriction: $\det(A - \lambda I_{n}) = 0$.

My concerns

I am mainly concerned with the wording of my proof. Could someone point out any theoretical flaw or missing step? Maybe I am overcomplicating things. Please let me know. Any contribution is appreciated.

Answer 1

Given that $\lambda$ is an eigenvalue of $A,$ there exists a nonzero vector $v$ such that $Av = \lambda v,$ from which it follows that $(A - \lambda I_n) v = Av - \lambda v = 0.$ But this says exactly that $\ker(A - \lambda I_n)$ is nontrivial (since it contains the nonzero vector $v$), hence $A - \lambda I_n$ is not invertible.

Conversely, if $\det(A - \lambda I_n) = 0$ for some scalar $\lambda,$ then there exists a nonzero vector $v$ in $\ker(A - \lambda I_n).$ By definition, we have that $0 = (A - \lambda I_n) v = Av - \lambda v$ so that $Av = \lambda v.$ QED.