[proofplan] Choose an ordered basis of $V$ and represent $T$ by a square matrix $A$. The [characteristic polynomial](/page/Characteristic%20Polynomial) $p(X)=\det(XI_n-A)$ has positive degree, so algebraic closedness gives a scalar $\lambda \in k$ with $p(\lambda)=0$. The determinant criterion then makes $\lambda I_n-A$ singular, hence it has a nonzero kernel vector. Translating that kernel vector back through the chosen basis gives a nonzero vector $v \in V$ satisfying $T(v)=\lambda v$. [/proofplan]

[step:Represent the linear map by a matrix in an ordered basis] Set \begin{align*} n:=\dim_k V. \end{align*} By hypothesis, $n \geq 1$. Choose an ordered basis \begin{align*} \mathcal B=(e_1,\dots,e_n) \end{align*} of the $k$-[vector space](/page/Vector%20Space) $V$. Let \begin{align*} \Phi:k^n \to V \end{align*} be the coordinate isomorphism defined by \begin{align*} \Phi(a_1,\dots,a_n)=\sum_{i=1}^n a_i e_i. \end{align*} Define the matrix $A=(A_{ij}) \in k^{n\times n}$ by the requirement that, for each $j \in \{1,\dots,n\}$, \begin{align*} T(e_j)=\sum_{i=1}^n A_{ij}e_i. \end{align*} Then $A$ represents $T$ in the basis $\mathcal B$, meaning that for every $x \in k^n$, \begin{align*} \Phi(Ax)=T(\Phi(x)). \end{align*} [/step]

[step:Choose a root of the characteristic polynomial]Let $I_n \in k^{n\times n}$ denote the identity matrix. Define the characteristic polynomial $p \in k[X]$ of the matrix $A$ by \begin{align*} p(X):=\det(XI_n-A). \end{align*} Since $A$ is an $n\times n$ matrix, the polynomial $p$ is monic of degree $n$. Because $n \geq 1$, the polynomial $p$ has positive degree. Since $k$ is algebraically closed, there exists $\lambda \in k$ such that \begin{align*} p(\lambda)=0. \end{align*} By the definition of $p$, this says \begin{align*} \det(\lambda I_n-A)=0. \end{align*}[/step]

[guided]The goal is to turn algebraic closedness into an eigenvector. Algebraic closedness applies to polynomials, so we attach to the matrix representative $A$ its characteristic polynomial. Let $I_n \in k^{n\times n}$ be the identity matrix, and define \begin{align*} p(X):=\det(XI_n-A) \in k[X]. \end{align*} This polynomial has degree $n$ and leading coefficient $1$: in the determinant expansion of $XI_n-A$, the product of the diagonal entries contributes $X^n$, and no other term has degree greater than $n$. Since $n=\dim_k V\geq 1$, the polynomial $p$ is nonconstant. Now we use exactly the hypothesis that $k$ is algebraically closed. A nonconstant polynomial over an [algebraically closed field](/page/Algebraically%20Closed%20Field) has a root in that field. Therefore there exists $\lambda \in k$ such that \begin{align*} p(\lambda)=0. \end{align*} Substituting the definition of $p$ gives \begin{align*} \det(\lambda I_n-A)=0. \end{align*} This scalar $\lambda$ is the candidate eigenvalue; the next step converts the determinant equation into a nonzero eigenvector.[/guided]

[step:Convert the singular matrix into a nonzero kernel vector] Set \begin{align*} B:=\lambda I_n-A \in k^{n\times n}. \end{align*} The determinant criterion for square matrices gives that $\det B=0$ if and only if the [linear map](/page/Linear%20Map) \begin{align*} B:k^n \to k^n \end{align*} is not invertible. Hence $B:k^n \to k^n$ is not invertible. Since $k^n$ is finite-dimensional and $B$ is an endomorphism of $k^n$, non-invertibility is equivalent to non-injectivity. Therefore there exists a vector $x \in k^n \setminus \{0\}$ such that \begin{align*} Bx=0. \end{align*} Substituting $B=\lambda I_n-A$, we obtain \begin{align*} (\lambda I_n-A)x=0. \end{align*} Equivalently, \begin{align*} Ax=\lambda x. \end{align*} [/step]

[step:Translate the matrix eigenvector back to the original vector space] Define \begin{align*} v:=\Phi(x) \in V. \end{align*} Since $\Phi:k^n \to V$ is an isomorphism and $x \neq 0$, we have $v \neq 0$. Using the matrix representation identity from the first step and the equality $Ax=\lambda x$, we compute \begin{align*} T(v)=T(\Phi(x)). \end{align*} The representation identity gives \begin{align*} T(\Phi(x))=\Phi(Ax). \end{align*} Since $Ax=\lambda x$ and $\Phi$ is $k$-linear, \begin{align*} \Phi(Ax)=\Phi(\lambda x)=\lambda \Phi(x)=\lambda v. \end{align*} Therefore \begin{align*} T(v)=\lambda v. \end{align*} We have found $\lambda \in k$ and $v \in V\setminus\{0\}$ satisfying the eigenvector equation, so $v$ is an eigenvector of $T$. [/step]