Objectives: Topics
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Coordinates relative to a basis.
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Dimension of a subspace.
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The rank of a matrix.
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Section 2.9 Dimension and Rank
Handout 2.9 Dimension and Rank
Objectives: Goals
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Calculate coordinates of a vector in a given basis.
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Characterize subspaces using the concept of dimension.
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Characterize matrices using rank, column space, and null space.
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Apply the Rank, Basis, and Invertibility Theorems to describe matrices and subspaces.
Key idea: Many different bases may be chosen for a given subspace. The choice can dramatically affect coordinate representations.
Example 2.57.
Sketch \(\vec b_1 + \vec b_2\) in each coordinate system below.
Two grids with different bases; vectors \(\vec b_1\) and \(\vec b_2\) are drawn.
Definition 2.58. Coordinates Relative to a Basis.
Let \(\mathcal{B}=\{\vec b_1,\dots,\vec b_p\}\) be a basis for a subspace \(H\text{.}\) For any \(\vec x\in H\text{,}\) the coordinates of \(\vec x\) relative to \(\mathcal B\) are the scalars \(c_1,\dots,c_p\) such that
\begin{equation*}
\vec x = c_1\vec b_1 + \cdots + c_p\vec b_p.
\end{equation*}
\begin{equation*}
[\vec x]_{\mathcal B} =
\begin{bmatrix}
c_1\\ \vdots\\ c_p
\end{bmatrix}.
\end{equation*}
Example 2.59.
Let:
\begin{equation*}
\vec v_1=\begin{bmatrix}1\\0\\1\end{bmatrix},\qquad
\vec v_2=\begin{bmatrix}1\\1\\1\end{bmatrix},\qquad
\vec x=\begin{bmatrix}5\\3\\5\end{bmatrix}.
\end{equation*}
Verify that \(\vec x\) is in the span of \(\mathcal B=\{\vec v_1,\vec v_2\}\text{,}\) and compute \([\vec x]_{\mathcal B}\text{.}\)
Example 2.60.
Let
\begin{equation*}
\vec v_1=\begin{bmatrix}-1\\0\\1\\0\end{bmatrix},\qquad
\vec v_2=\begin{bmatrix}0\\1\\3\\2\end{bmatrix},\qquad
\vec v_3=\begin{bmatrix}3\\6\\2\\-1\end{bmatrix},\qquad
\vec x=\begin{bmatrix}3\\11\\4\\-4\end{bmatrix},
\end{equation*}
and \(H = \textrm{Span}\{\vec v_1, \vec v_2, \vec v_3\}\text{.}\)
(a)
(b)
Determine if \(\vec x \in H\text{.}\)
(c)
Compute \([\vec x]_{\mathcal B}\text{.}\)
Definition 2.61. Dimension.
The dimension of a nonzero subspace \(H\text{,}\) written \(\operatorname{dim}(H)\text{,}\) is the number of vectors in any basis of \(H\text{.}\) We define \(\operatorname{dim}(\{0\})=0\text{.}\)
Theorem 2.62. Dimension Is Well-Defined.
Any two bases \(\mathcal B_1\) and \(\mathcal B_2\) of a nonzero subspace \(H\) have the same number of vectors.
Note 2.63.
Some Examples:-
\(\operatorname{dim}(\mathbb{R}^n)=n\text{.}\)
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\(H=\{(x_1,\dots,x_n): x_1+\cdots+x_n=0\}\) has dimension \(n-1\text{.}\)
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\(\operatorname{dim}(\operatorname{Nul}A)\) is the number of free variables.
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\(\operatorname{dim}(\operatorname{Col}A)\) is the number of pivotal columns.
Definition 2.64. Rank.
The rank of a matrix \(A\) is the dimension of its column space.
Example 2.65.
Compute \(\operatorname{rank}(A)\) and \(\operatorname{dim}(\operatorname{Nul}A)\) and find a basis for \(\Col(A)\) and \(\Nul(A)\text{,}\) where
\begin{equation*}
A = \begin{bmatrix}
2 \amp -3 \amp -12 \amp 4 \amp -7\\
4 \amp -5 \amp -22 \amp 3 \amp 1\\
-5 \amp 2 \amp 19 \amp -10 \amp -10\\
0 \amp 2 \amp 4 \amp 7 \amp -4
\end{bmatrix}
\sim
\begin{bmatrix}
1 \amp 0 \amp -3 \amp 0 \amp 8\\
0 \amp 1 \amp 2 \amp 0 \amp 5\\
0 \amp 0 \amp 0 \amp 1 \amp -2\\
0 \amp 0 \amp 0 \amp 0 \amp 0
\end{bmatrix}.
\end{equation*}
Theorem 2.66. Rank Theorem.
\begin{equation*}
\operatorname{rank}(A)+\operatorname{dim}(\operatorname{Null}A)=n.
\end{equation*}
Theorem 2.67. Basis Theorem.
Any two bases for a subspace have the same number of vectors. That common number of vectors is the dimension of the subspace.
Theorem 2.68. The Invertible Matrix Theorem (Continued).
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(a) \(A\) is invertible.
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(m) The columns of \(A\) form a basis for \(\mathbb{R}^n\text{.}\)
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(n) \(\operatorname{Col}A=\mathbb{R}^n\text{.}\)
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(o) \(\operatorname{dim}\operatorname{Col}A=n\text{.}\)
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(p) \(\operatorname{rank}A=n\text{.}\)
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(q) \(\operatorname{Nul}A=\{\vec 0\}\text{.}\)
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(r) \(\operatorname{dim}\operatorname{Nul}A=0\text{.}\)
Example 2.69.
For each part below, give an example of a \(2\times 3\) matrix \(A\) in RREF with the stated property, if possible.
(a)
\(\operatorname{rank}(A)=3\)
(b)
\(\operatorname{rank}(A)=2\)
(c)
\(\operatorname{dim}(\operatorname{Null}A)=2\)
(d)
\(\operatorname{Nul}(A)=\{\vec 0\}\)
