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Let $Z$ be a countable set. Let $f_1,....,f_n$ be a collection of real functions over $Z$. Let $z_1,...,z_m,...$ be an enumeration of elements of $Z$.

Define $V_1 = \{ (f_1(z),...,f_n(z)) | z \in Z\}$ a set of $n$ dimensional vectors. Define $V_2 =\{ (f_i(z_1),...,f_i(z_m),..., )) | i \in\{ 1,...,n\} \}$ a set of infinite dimensional vectors.

Let $span(V_i)$ be the span (linear span) of $V_i$. Is the linear dimension of $span(V_i)$ the same for $i=1,2$?

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+1 for not assuming that the answer is obviously yes. – Qiaochu Yuan Oct 25 '10 at 15:55
up vote 3 down vote accepted

Let me rephrase your question in a way that's less confusing. Define a matrix $A$ such that $A_{ij} = f_i(z_j)$. This matrix has finitely many rows but infinitely many columns. Then the dimension of the span of $V_1$ is the row rank of this matrix, and the dimension of the span of $V_2$ is the column rank of this matrix. And you want to know if they are the same.

The answer is yes. Here's one proof: define $A_m$ to be $A$, but with all columns after the $m^{th}$ column chopped off, and let $a_m$ denote the column rank of $A_m$. Then it's not hard to see that there exists $m$ such that $a_m$ is the column rank of $A$. Now apply the theorem that row rank equals column rank for ordinary matrices to $A_m$.

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