[proofplan] The statement is a biconditional relating two a priori different conditions on a scalar $\lambda$: the existence of a nonzero vector $v$ with $Av = \lambda v$ (the eigenvalue condition), and the vanishing of the polynomial $\chi_A(t) = \det(A - tI)$ at $t = \lambda$ (the algebraic condition). We prove both directions simultaneously by reformulating the eigenvalue condition as the existence of a nonzero kernel of $A - \lambda I$, then invoking the standard chain of equivalences linking nonzero kernel, non-invertibility, and vanishing determinant for a square matrix. Each link is a named characterisation that applies to every square matrix over the scalar field, so the chain produces a genuine "if and only if". [/proofplan] [step:Rewrite the eigenvalue condition as a kernel condition on $A - \lambda I$] By definition, $\lambda$ is an eigenvalue of $A$ if and only if there exists a nonzero vector $v \in \mathbb{R}^n$ such that $Av = \lambda v$. Subtracting $\lambda v = \lambda I v$ from both sides and collecting yields the equivalent condition \begin{align*} (A - \lambda I)v &= \mathbf{0}, \qquad v \neq \mathbf{0}. \end{align*} The existence of such a $v$ is precisely the statement that the kernel of the [linear map](/page/Linear%20Map) $A - \lambda I : \mathbb{R}^n \to \mathbb{R}^n$ is nontrivial: \begin{align*} \lambda \text{ is an eigenvalue of } A \quad &\iff \quad \ker(A - \lambda I) \neq \{\mathbf{0}\}. \end{align*} [guided] We begin from the [definition of an eigenvalue](/page/Eigenvalue): $\lambda$ is an eigenvalue of $A$ exactly when there is a nonzero $v \in \mathbb{R}^n$ satisfying $Av = \lambda v$. The nonzero requirement is essential — the equation $Av = \lambda v$ is satisfied by $v = \mathbf{0}$ for every $\lambda$, so without it the notion of eigenvalue would be meaningless. Our goal is to connect this condition to the characteristic polynomial, whose definition involves the matrix $A - \lambda I$, not $A$ directly. The bridge is the elementary rewrite \begin{align*} Av = \lambda v \quad \iff \quad Av - \lambda v = \mathbf{0} \quad \iff \quad (A - \lambda I)v = \mathbf{0}, \end{align*} where the second step uses $\lambda v = (\lambda I)v$ to factor the scalar through the identity matrix. This is the key algebraic move: the single matrix $A - \lambda I$ now carries all information about whether $\lambda$ is an eigenvalue. Translating "there exists a nonzero solution of $(A - \lambda I)v = \mathbf{0}$" into kernel language: the kernel of the linear map $A - \lambda I : \mathbb{R}^n \to \mathbb{R}^n$ is the set of all $v$ with $(A - \lambda I)v = \mathbf{0}$. The zero vector always lies in the kernel, so asking for a nonzero solution is asking that the kernel be strictly larger than $\{\mathbf{0}\}$. Hence \begin{align*} \lambda \text{ is an eigenvalue of } A \quad \iff \quad \ker(A - \lambda I) \neq \{\mathbf{0}\}. \end{align*} Why is this reformulation useful? Because "nontrivial kernel" is the standard entry point for the invertibility criterion in the next step — once we phrase the eigenvalue condition as a statement about $A - \lambda I$ having a nonzero kernel, we can apply general theorems about square matrices to replace it with a determinant condition. [/guided] [/step] [step:Chain the standard equivalences linking nontrivial kernel, non-invertibility, and vanishing determinant] The matrix $A - \lambda I$ is a square $n \times n$ matrix. Two standard characterisations apply: (i) The invertibility criterion for square matrices states that for $M \in \mathbb{R}^{n \times n}$, the following are equivalent: $M$ is invertible; $\ker(M) = \{\mathbf{0}\}$; the columns of $M$ are linearly independent; the rows of $M$ are linearly independent. Taking the contrapositive of the equivalence "$M$ invertible $\iff$ $\ker(M) = \{\mathbf{0}\}$" with $M := A - \lambda I$ gives \begin{align*} \ker(A - \lambda I) \neq \{\mathbf{0}\} \quad \iff \quad A - \lambda I \text{ is not invertible}. \end{align*} (ii) The [determinantal criterion for invertibility](/theorems/396) states that for any square matrix $M \in \mathbb{R}^{n \times n}$, $M$ is invertible if and only if $\det(M) \neq 0$. Again taking the contrapositive with $M := A - \lambda I$: \begin{align*} A - \lambda I \text{ is not invertible} \quad \iff \quad \det(A - \lambda I) = 0. \end{align*} Concatenating the two biconditionals with the one from the previous step, and recognising $\det(A - \lambda I) = \chi_A(\lambda)$ by the definition of the characteristic polynomial, we obtain \begin{align*} \lambda \text{ is an eigenvalue of } A \quad &\iff \quad \ker(A - \lambda I) \neq \{\mathbf{0}\} \\ &\iff \quad A - \lambda I \text{ is not invertible} \\ &\iff \quad \det(A - \lambda I) = 0 \\ &\iff \quad \chi_A(\lambda) = 0, \end{align*} which is the claim. [guided] From the previous step we have reduced the eigenvalue condition to a statement about the kernel of $A - \lambda I$. To reach the characteristic polynomial, we need two further equivalences. Both are general facts about a single square matrix $M \in \mathbb{R}^{n \times n}$; we will apply them with $M = A - \lambda I$. **First equivalence: nontrivial kernel $\iff$ non-invertibility.** The invertibility criterion for square matrices asserts that for $M \in \mathbb{R}^{n \times n}$, invertibility is equivalent to injectivity as a linear map, which in turn is equivalent to $\ker(M) = \{\mathbf{0}\}$. Why does squareness matter? Because for non-square matrices injectivity and surjectivity decouple, and "invertible" no longer has a single unambiguous meaning; for square matrices, the [rank-nullity theorem](/theorems/916) collapses these notions to one. Applying the contrapositive with $M := A - \lambda I$ (which is square because $A$ is $n \times n$): \begin{align*} \ker(A - \lambda I) \neq \{\mathbf{0}\} \quad \iff \quad A - \lambda I \text{ is not invertible}. \end{align*} **Second equivalence: non-invertibility $\iff$ vanishing determinant.** The [determinantal criterion](/theorems/396) states that a square matrix $M$ is invertible if and only if $\det(M) \neq 0$. This is again a general fact — one direction follows from the multiplicativity of the determinant (if $M M^{-1} = I$ then $\det(M)\det(M^{-1}) = 1$, so $\det(M) \neq 0$), and the converse is supplied by the adjugate formula $M^{-1} = \frac{1}{\det M}\operatorname{adj}(M)$, valid whenever $\det M \neq 0$. Contrapositive with $M := A - \lambda I$: \begin{align*} A - \lambda I \text{ is not invertible} \quad \iff \quad \det(A - \lambda I) = 0. \end{align*} **Closing the chain.** The [characteristic polynomial](/page/Characteristic%20Polynomial) of $A$ is defined by $\chi_A(t) := \det(A - tI)$, so evaluating at $t = \lambda$ gives $\chi_A(\lambda) = \det(A - \lambda I)$. This is a definitional identity, not a theorem — no hypotheses to verify. Stringing together the four equivalences: \begin{align*} \lambda \text{ is an eigenvalue of } A \quad &\iff \quad \ker(A - \lambda I) \neq \{\mathbf{0}\} \quad &&\text{(Step 1)} \\ &\iff \quad A - \lambda I \text{ is not invertible} \quad &&\text{(invertibility criterion)} \\ &\iff \quad \det(A - \lambda I) = 0 \quad &&\text{(determinantal criterion)} \\ &\iff \quad \chi_A(\lambda) = 0 \quad &&\text{(definition of } \chi_A\text{)}. \end{align*} Each link is a true biconditional, so the composition is a biconditional, which proves the theorem. Every equivalence held for both directions simultaneously, so there is no asymmetry between "$\lambda$ eigenvalue $\implies$ $\chi_A(\lambda) = 0$" and its converse — both follow from the same chain, read forwards or backwards. [/guided] [/step]