Definition 6.4.1.
The dimension of a non-zero subspace \(H\text{,}\) denoted by \(\dim{H}\text{,}\) is the number of vectors in any basis of \(H\text{.}\) The dimension of the zero subspace \(\set{0}\) is defined to be 0.
Section 6.4 Dimension and Rank
Recall that bases of a subspace are not unique. However, it turns out that all bases of a given subspace contain the same number of vectors. That is, if a subspace \(H\) has a basis consisting of \(p\) vectors, then every basis of \(H\) contains exactly \(p\) vectors. This number \(p\) is called the dimension of \(H\text{.}\)
Subsection 6.4.1 Dimension of a Subspace
Example 6.4.2. The space \(\mathbb{R}^n\).
Intuitively, the vector space \(\mathbb{R}^n\) has dimension \(n\text{.}\) Recall that the standard basis of \(\mathbb{R}^n\) includes \(n\) vectors, so any basis of \(\mathbb{R}^n\) has \(n\) vectors.
Example 6.4.3. Dimension of the null space.
The dimension of the null space of a matrix \(A\) is, by definition, equal to the number of vectors in its basis. Recall that there is one vector for every free variable, and so the dimension is equal to the number of free variables when solving \(A\vec{x} = \vec{0}\text{.}\)
Subsection 6.4.2 Rank of a Matrix
Definition 6.4.4.
The rank of a matrix \(A\text{,}\) denoted by \(\rank{A}\text{,}\) is the dimension of the column space of \(A\text{.}\) In other words,
\begin{equation*}
\rank{A} = \dim{\Span{\vec{a}_1, \dots, \vec{a}_m}}
\end{equation*}
Then, the rank is also the number of pivot columns (recall that the dimension of \(C(A)\) is equal to the number of pivot columns).
Example 6.4.5. Dimension of the column space.
Consider the dimension of the column space of a matrix. Recall that the pivot columns of \(A\) form a basis of \(A\text{.}\) Thus, the dimension of \(A\) is the number of pivot columns of \(A\text{.}\)
Together, the dimension of the column space is the number of pivot columns, and the dimension of the null space is the number of non-pivot columns. For an \(n \times m\) matrix, there are \(n\) columns in total. Thus, in summary,
Theorem 6.4.6. Rank theorem.
Let \(A\) be a matrix with \(n\) columns. Then,
\begin{equation*}
\boxed{\rank{A} + \dim{N(A)} = n}
\end{equation*}
In particular, if say \(\rank{A} = r\text{,}\) then \(\dim{N(A)} = n - r\text{.}\)
Theorem 6.4.7. The basis theorem.
Let \(H\) be a \(p\)-dimensional subspace of \(\mathbb{R}^n\) (i.e. \(\dim{H} = p\)). Then, every linearly independent set of \(p\) vectors is a basis of \(H\text{.}\) Also, every set of \(p\) vectors that spans \(H\) is a basis of \(H\text{.}\)
Subsection 6.4.3 Rank and the Invertible Matrix Theorem
Theorem 6.4.8.
Let \(A\) be an \(n \times n\) matrix. Then, the following are each equivalent to the statement that \(A\) is invertible:
- The columns of \(A\) form a basis of \(\mathbb{R}^n\text{.}\)
- The columns of \(A\) span \(\mathbb{R}^n\text{,}\) \(\Col{A} = \mathbb{R}^n\text{.}\)
- \(\dim{\Col{A}} = n\text{.}\)
- \(A\) has full rank, \(\rank{A} = n\text{.}\)
- \(N(A) = \set{\vec{0}}\text{.}\)
- \(\dim{N(A)} = 0\text{.}\)
