22 - Dimension

Dimension

Let $V$ be a finite-dimensional vector space
The dimension of $V$ written as $\dim V$ is the size of any basis of $V$

Row Rank (using Dimension)

Row Rank of a matrix is the dimension of the row space
Also the number of non-zero rows of RREF of the matrix

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Dimension of Subspaces

Let $U$ be a subspace of a finite-dimensional vector space $V$
Then $U$ is finite-dimensional and

$$\dim U=\dim U$$

However if $\dim U=\dim V$ then

$$U=V$$

Proof

Let $n=\dim V$
By 05 - Size Inequality for Independent vs Spanning Sets then each linearly independent subset of $V$ has size at most $n$
Let $S$ be a largest linearly independent set contained in $U$ ($S\subseteq U\subseteq V$) hence

$$|S|\le n$$

Suppose for a contradiction that $⟨S⟩\ne U$ then there exists $u\in U\setminus ⟨S⟩$
By Extension of a Linearly Independent Set then $S\cup \{u\}$ is linearly independent and

$$|S\cup \{u\}|>|S|$$

This is a contradiction for the definition of $S$ so $U=⟨S⟩$
As $S$ is linearly independent then $S$ is a basis of $U$ so

$$|S|\le n$$

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Suppose $\dim U=\dim V$ and $U\ne V$
Then there exists $v\in V\setminus U$
Hence $v$ may be added to the basis of $U$ to create a linearly independent subset of $V$

$$\dim U+1=\dim V+1$$

Which is a contradiction so $\dim U=\dim V$ implies $U=V$