Cardinality of Linearly Independent Set is No Greater than Dimension

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Theorem

Let $E$ be a vector space of $n$ dimensions.

Let $H$ be a linearly independent subset of $E$.


Then:

$H$ has at most $n$ elements.


Proof

Let $H$ be a linearly independent subset of $E$.

By definition of dimension of vector space, $E$ has a basis with exactly $n$ elements.

By Sufficient Conditions for Basis of Finite Dimensional Vector Space, $B$ is a generator for $E$.

Then by Size of Linearly Independent Subset is at Most Size of Finite Generator, $H$ has at most $n$ elements.

$\blacksquare$


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