In the Wikipedia article on Hilbert's Nullstensatz, http://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz the following application of the Weak Nullstensatz is mentioned:

Commuting matrices

The fact that commuting matrices have a common eigenvector – and hence by induction stabilize a common flag and are simultaneously triangularizable – can be interpreted as a result of the weak Nullstellensatz, as follows: commuting matrices form a commutative algebra $K[A_1,\ldots,A_k]$ over $K[x_1,\ldots,x_k];$ the matrices satisfy various polynomials such as their minimal polynomials, which form a proper ideal (because they are not all zero, in which case the result is trivial); one might call this the '''characteristic ideal''', by analogy with the characteristic polynomial.

One then defines an eigenvector for a commutative algebra as a vector $v$ such that for all $x \in A$ one has $x(v) = \lambda(x)\cdot v$ for a linear functional $\lambda\colon A \to K.$ This simply linearizes the definition of an eigenvalue, and is the correct definition for a common eigenvector, as if $v$ is a common eigenvector, meaning $A_i(v)=\lambda_i v,$ then the functional is defined as $\textstyle{\lambda(c_0 I + c_1 A_1 + \cdots c_k A_k) := c_0 + \sum c_i \lambda_i}$ (treating scalars as multiples of the identity matrix $A_0 := I$, which has eigenvalue 1 for all vectors), and conversely an eigenvector for such a functional $\lambda$ is a common eigenvector. Geometrically, the eigenvalue corresponds to the point in affine $k$-space with coordinates $(\lambda_1,\ldots,\lambda_k)$ with respect to the basis given by $A_i.$

Then the existence of an eigenvalue $\lambda$ is equivalent to the ideal generated by (the relations satisfied by) $A_i$ being non-empty, which exactly generalizes the usual proof of existence of an eigenvalue existing for a single matrix over an algebraically closed field by showing that the characteristic polynomial has a zero.

I am somewhat struggling to make sense of that. The Weak Nullstellensatz says that I find a functional $\lambda \colon K[A_1, \dots, A_k] \to K$, and $\lambda(A_i)$ is an eigenvalue of $A_i$. But how do I conclude that a common eigenvector exists?