[proofplan] We unfold the definition of eigenvalue and translate the equation $Av=\lambda v$ into a kernel condition for the matrix $\lambda I_n-A$. The determinant criterion for singularity then converts the existence of a nonzero kernel vector into the vanishing of $\det(\lambda I_n-A)$. Finally, by the defining convention for the [characteristic polynomial](/page/Characteristic%20Polynomial), this determinant is exactly $\chi_A(\lambda)$. [/proofplan]

[step:Translate the eigenvalue condition into a kernel condition] Let $L_\lambda:k^n\to k^n$ denote the $k$-[linear map](/page/Linear%20Map) represented in the standard basis by the matrix $\lambda I_n-A$. By definition, $\lambda$ is an eigenvalue of $A$ if and only if there exists a vector $v\in k^n$ with $v\ne 0$ such that $Av=\lambda v$. Subtracting $Av$ from both sides gives $(\lambda I_n-A)v=0$. Thus $\lambda$ is an eigenvalue of $A$ if and only if $\ker L_\lambda$ contains a nonzero vector. [/step]

[step:Convert the nonzero kernel condition into determinant vanishing]The determinant criterion for an $n\times n$ matrix over a field says that a matrix $B\in M_n(k)$ has a nonzero vector in its kernel if and only if $\det B=0$. Applying this criterion to $B=\lambda I_n-A$ gives $\ker L_\lambda\ne \{0\}$ if and only if $\det(\lambda I_n-A)=0$.[/step]

[guided]We now need to turn the equation \begin{align*} (\lambda I_n-A)v=0 \end{align*} for some nonzero $v\in k^n$ into a scalar equation. The bridge is the determinant criterion for square matrices over a field: for a matrix $B\in M_n(k)$, the following are equivalent: $B$ has a nonzero kernel vector, the associated linear map $k^n\to k^n$ is not injective, the associated linear map is not invertible, and $\det B=0$. Here the relevant matrix is \begin{align*} B=\lambda I_n-A. \end{align*} The linear map represented by this matrix is exactly $L_\lambda:k^n\to k^n$. Therefore $L_\lambda$ has a nonzero kernel vector if and only if \begin{align*} \det(\lambda I_n-A)=0. \end{align*} This is precisely the point of passing from $Av=\lambda v$ to $(\lambda I_n-A)v=0$: it places the eigenvalue condition into the kernel of a square matrix, where the determinant detects singularity.[/guided]

[step:Identify the determinant with the value of the characteristic polynomial] By the convention in the statement, $\chi_A(t)=\det(tI_n-A)$. Evaluating this polynomial at $t=\lambda$ gives $\chi_A(\lambda)=\det(\lambda I_n-A)$. Combining this identity with the previous two steps, $\lambda$ is an eigenvalue of $A$ if and only if $\chi_A(\lambda)=0$. This proves the theorem. [/step]