(I) Coordinate Systems
Suppose the set \( \mathcal{B} = \{\vec{b}_1, \dots, \vec{b}_p\} \) is a basis for a subspace \(H\). Each \( \vec{x} \) in \(H\) can be written in only one way as a linear combination of the basis vectors.
Definition: Coordinates
Suppose the set \( \mathcal{B} = \{\vec{b}_1, \dots, \vec{b}_p\} \) is a basis for a subspace \(H\). For each \(\vec{x}\) in \(H\), the coordinates of \(\vec{x}\) relative to \(\mathcal{B}\) are the weights \(c_1, \dots, c_p\) such that:
The coordinate vector of \(\vec{x}\) relative to \(\mathcal{B}\) (or the \(\mathcal{B}\)-coordinate vector of \(\vec{x}\)) is denoted as:
Let \( \vec{b}_1 = \begin{bmatrix} -3 \\ 1 \\ 4 \end{bmatrix} \), \( \vec{b}_2 = \begin{bmatrix} 7 \\ 5 \\ -6 \end{bmatrix} \), and \( \vec{x} = \begin{bmatrix} 11 \\ 0 \\ -13 \end{bmatrix} \).
The set \( \mathcal{B} = \{\vec{b}_1, \vec{b}_2\} \) is a basis for \(H = \text{Span}\{\vec{b}_1, \vec{b}_2\}\). Determine if \(\vec{x}\) is in \(H\) and if it is, find \([\vec{x}]_{\mathcal{B}}\).
Solution:
We must find weights \(c_1\) and \(c_2\) such that \( c_1\vec{b}_1 + c_2\vec{b}_2 = \vec{x} \).
Set up the augmented matrix and reduce it:
\[ \begin{bmatrix} -3 & 7 & 11 \\ 1 & 5 & 0 \\ 4 & -6 & -13 \end{bmatrix} \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & 22 & 11 \\ 0 & -26 & -13 \end{bmatrix} \sim \begin{bmatrix} 1 & 5 & 0 \\ 0 & 1 & 1/2 \\ 0 & 0 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -5/2 \\ 0 & 1 & 1/2 \\ 0 & 0 & 0 \end{bmatrix} \]Because the system is consistent, \(\vec{x}\) is in \(H\). The weights are \(c_1 = -5/2\) and \(c_2 = 1/2\).
Therefore, the coordinate vector is:
\[ [\vec{x}]_{\mathcal{B}} = \begin{bmatrix} -5/2 \\ 1/2 \end{bmatrix} \]Geometric Interpretation:
Points in \(H\) are also in \(\mathbb{R}^3\). But they are completely determined by their coordinate vectors \([\vec{x}]_{\mathcal{B}}\), which are in \(\mathbb{R}^2\). In a sense, \(H\) "looks like" \(\mathbb{R}^2\).
The mapping \(\vec{x} \mapsto [\vec{x}]_{\mathcal{B}}\) is a one-to-one correspondence between \(H\) and \(\mathbb{R}^2\) that preserves linear combinations. We say that \(H\) is isomorphic to \(\mathbb{R}^2\) (even though \(H \neq \mathbb{R}^2\)).
In general, if \(\mathcal{B} = \{\vec{b}_1, \dots, \vec{b}_p\}\) is a basis for \(H\), then \(H\) is isomorphic to \(\mathbb{R}^p\).
(II) The Dimension of a Subspace
It can be shown that if a subspace \(H\) has a basis of \(p\) vectors, then every basis of \(H\) must consist of exactly \(p\) vectors. This makes the concept of dimension well-defined.
Definitions
Dimension: The dimension of a nonzero subspace \(H\), denoted \(\dim H\), is the number of vectors in any basis for \(H\).
Note: The dimension of the zero subspace \(\{\vec{0}\}\) is defined to be 0 (since it has 0 basis vectors, as \(\{\vec{0}\}\) itself is a linearly dependent set).
Rank: The rank of a matrix \(A\), denoted \(\text{rank } A\), is the dimension of the column space of \(A\) (\(\text{Col } A\)).
(This is true because the pivot columns of \(A\) form a basis for \(\text{Col } A\)).
Determine \(\text{rank } A\) and \(\dim \text{Nul } A\) for a matrix \(A\) that row reduces to the following echelon form:
\[ A \sim \begin{bmatrix} 1 & -2 & 9 & 5 & 4 \\ 0 & 1 & -3 & 0 & -7 \\ 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]Solution:
Notice there are 3 pivot positions (columns 1, 2, and 4). Therefore, \(\text{rank } A = 3\).
There are 2 columns without pivots (columns 3 and 5), which correspond to 2 free variables. Therefore, \(\dim \text{Nul } A = 2\).
Theorem 14 (The Rank Theorem)
If a matrix \(A\) has \(n\) columns, then:
(Number of pivot columns + Number of non-pivot columns = Total number of columns).
Theorem 15 (The Basis Theorem)
Let \(H\) be a \(p\)-dimensional subspace of \(\mathbb{R}^n\).
- Any linearly independent set of exactly \(p\) elements in \(H\) is automatically a basis for \(H\).
- Any set of exactly \(p\) elements of \(H\) that spans \(H\) is automatically a basis for \(H\).
The Invertible Matrix Theorem (Continued)
Let \(A\) be an \(n \times n\) matrix. The following statements are equivalent to \(A\) being an invertible matrix:
- The columns of \(A\) form a basis of \(\mathbb{R}^n\).
- \(\text{Col } A = \mathbb{R}^n\).
- \(\dim \text{Col } A = n\).
- \(\text{rank } A = n\).
- \(\text{Nul } A = \{\vec{0}\}\).
- \(\dim \text{Nul } A = 0\).
(III) Examples and Applications
Let \(A\) be a \(5 \times 7\) matrix and \(\text{rank } A = 5\).
(1) Is \(\text{Col } A = \mathbb{R}^5\)?
Solution: Yes. \(\text{Col } A\) is a subspace of \(\mathbb{R}^5\), and \(\dim \text{Col } A = 5\). Since it has the same dimension as the entire space \(\mathbb{R}^5\), \(\text{Col } A = \mathbb{R}^5\).
(2) Is \(\text{Nul } A = \mathbb{R}^2\)?
Solution: No. By the Rank Theorem, \(\dim \text{Nul } A = 7 - 5 = 2\). However, \(\text{Nul } A\) is a subspace of \(\mathbb{R}^7\) (the domain), not \(\mathbb{R}^2\). So \(\text{Nul } A \neq \mathbb{R}^2\) even though its dimension is 2.
Find a basis for the subspace spanned by the vectors:
Solution:
Let \( A = \begin{bmatrix} \vec{u}_1 & \vec{u}_2 & \vec{u}_3 & \vec{u}_4 & \vec{u}_5 \end{bmatrix} \). The subspace is \(\text{Col } A\).
Suppose row reducing \(A\) reveals pivot positions in columns 1, 2, and 4:
The pivot columns of the original matrix \(A\) form a basis for \(\text{Col } A\). Therefore, \(\{\vec{u}_1, \vec{u}_2, \vec{u}_4\}\) forms a basis for \(\text{Span}\{\vec{u}_1, \dots, \vec{u}_5\}\).
Note: We can also find other bases for \(\text{Col } A\). For example, \(\{\vec{u}_1, \vec{u}_3, \vec{u}_5\}\), \(\{\vec{u}_1, \vec{u}_2, \vec{u}_5\}\), or \(\{\vec{u}_2, \vec{u}_3, \vec{u}_4\}\). Since \(\dim \text{Col } A = 3\), any 3 linearly independent vectors in \(\text{Col } A\) will form a basis for \(\text{Col } A\).
Suppose columns 1, 3, 5, and 6 of a matrix \(A\) are linearly independent (but not necessarily pivot columns) and the rank of \(A\) is 4. Explain why the four columns mentioned must be a basis for \(\text{Col } A\).
Solution:
The columns \(\vec{a}_1, \vec{a}_3, \vec{a}_5\), and \(\vec{a}_6\) of \(A\) are linearly independent. Since \(\text{rank } A = \dim \text{Col } A = 4\), the set \(\{\vec{a}_1, \vec{a}_3, \vec{a}_5, \vec{a}_6\}\) is a linearly independent set of exactly 4 elements in a 4-dimensional column space.
By the Basis Theorem, any such set is automatically a basis for \(\text{Col } A\).
Let \(A\) be an \(m \times n\) matrix. Which of the following statements must be true?
(1) If the equation \(A\vec{x} = \vec{b}\) is consistent for each \(\vec{b} \in \mathbb{R}^m\), then \(\text{rank } A = n\).
Answer: False. If the equation is consistent for all \(\vec{b}\) in \(\mathbb{R}^m\), it means the columns of \(A\) span \(\mathbb{R}^m\). Therefore, \(A\) has a pivot in every row, which implies \(\text{rank } A = m\), not necessarily \(n\).
(2) If the equation \(A\vec{x} = \vec{0}\) has only the trivial solution, then \(\text{rank } A = n\).
Answer: True. Having only the trivial solution means the columns of \(A\) are linearly independent. Thus, every column of \(A\) is a pivot column, which implies \(\text{rank } A = n\).
(3) If \(\text{rank } A = n\) and \(A\) is a square matrix, then \(A\) is invertible.
Answer: True. This follows directly from the Invertible Matrix Theorem.
(4) If \(\text{rank } A = m\) and the linear transformation \(\vec{x} \mapsto A\vec{x}\) is one-to-one, then \(A\) is invertible.
Answer: True. If the transformation is one-to-one, the columns of \(A\) are linearly independent, meaning \(\text{rank } A = n\). We are given that \(\text{rank } A = m\). Therefore, \(m = n\), making \(A\) a square \(n \times n\) matrix with \(n\) pivots. By the Invertible Matrix Theorem, \(A\) is invertible.