answer
To show that \( A \) and \( Q \) have the same column space, we'll use the given hints:
Step 1: Show that \( \text{Col } A \subseteq \text{Col } Q \)
Given \( y \in \text{Col } A \), we can write \( y = Ax \) for some vector \( x \).
Since \( A = QR \) and \( R \) is invertible, we have:
\ y = Ax = QRx \\
Let \( x' = Rx \). Since \( R \) is invertible, \( x' \) can be any vector in \(\mathbb{R}^n\). Thus:
\ y = Qx' \\
This shows that \( y \) is also in the column space of \( Q \), i.e., \( \text{Col } A \subseteq \text{Col } Q \).
Step 2: Show that \( \text{Col } Q \subseteq \text{Col } A \)
Given \( y \in \text{Col } Q \), we can write \( y = Qx \) for some vector \( x \).
Since \( A = QR \) and \( R \) is invertible, we can multiply both sides by \( R^{-1} \):
\ Q = AR\^{-1} \\
Thus:
\ y = Qx = AR\^{-1}x \\
Let \( x' = R^{-1}x \). Since \( R \) is invertible, \( x' \) can be any vector in \(\mathbb{R}^n\). Thus:
\ y = Ax' \\
This shows that \( y \) is also in the column space of \( A \), i.e., \( \text{Col } Q \subseteq \text{Col } A \).
Conclusion
Since we've shown that \( \text{Col } A \subseteq \text{Col } Q \) and \( \text{Col } Q \subseteq \text{Col } A \), we conclude that \( \text{Col } A = \text{Col } Q \).
Therefore, \( A \) and \( Q \) have the same column space.
details
To understand why \( x' \) can be any vector in \(\mathbb{R}^n\), let's break it down step by step.
Understanding the Concept
Given that \( R \) is an invertible matrix:
- **Invertibility of \( R \)**: Since \( R \) is invertible, there exists a matrix \( R^{-1} \) such that:
\ R R\^{-1} = I \\
where \( I \) is the identity matrix.
- **Transformation by \( R \)**: Any vector \( x \in \mathbb{R}^n \) can be transformed by multiplying it by \( R \), resulting in a new vector \( x' \). We write:
\ x' = Rx \\
- **Invertible Transformation**: Because \( R \) is invertible, we can always find \( x \) given \( x' \) by multiplying \( x' \) by \( R^{-1} \):
\ x = R\^{-1} x' \\
Why \( x' \) Can Be Any Vector in \(\mathbb{R}^n\)
Let's see why \( x' \) can be any vector in \(\mathbb{R}^n\):
-
**Surjectivity**: The function \( f: \mathbb{R}^n \to \mathbb{R}^n \) defined by \( f(x) = Rx \) is surjective because for any vector \( x' \) in the output space (the codomain), there exists an \( x \) in the input space (the domain) such that \( Rx = x' \). This is because \( R \) has full rank (since it's invertible).
-
**Vector Space Mapping**: Since \( R \) maps \( \mathbb{R}^n \) to itself in a bijective manner (one-to-one and onto), for every vector \( x' \in \mathbb{R}^n \), there exists a vector \( x \in \mathbb{R}^n \) such that \( x' = Rx \).
-
**Existence of \( x \)**: Given any vector \( x' \in \mathbb{R}^n \), we can always find a corresponding vector \( x \) using \( x = R^{-1} x' \). Therefore, \( x' \) can be any vector in \(\mathbb{R}^n\).
Example
Let's consider a concrete example with \( R \):
- Suppose \( R \) is a 2x2 invertible matrix:
\ R = \\begin{pmatrix} 2 \& 1 \\\\ 1 \& 3 \\end{pmatrix} \\
- We can find \( R^{-1} \):
\ R\^{-1} = \\frac{1}{5} \\begin{pmatrix} 3 \& -1 \\\\ -1 \& 2 \\end{pmatrix} \\
- Now, given any vector \( x' \in \mathbb{R}^2 \), say \( x' = \begin{pmatrix} 4 \\ 5 \end{pmatrix} \):
\ x = R\^{-1} x' = \\frac{1}{5} \\begin{pmatrix} 3 \& -1 \\\\ -1 \& 2 \\end{pmatrix} \\begin{pmatrix} 4 \\\\ 5 \\end{pmatrix} = \\frac{1}{5} \\begin{pmatrix} 3 \\cdot 4 + (-1) \\cdot 5 \\\\ (-1) \\cdot 4 + 2 \\cdot 5 \\end{pmatrix} = \\frac{1}{5} \\begin{pmatrix} 12 - 5 \\\\ -4 + 10 \\end{pmatrix} = \\begin{pmatrix} 1.4 \\\\ 1.2 \\end{pmatrix} \\
This shows that for any \( x' \), we can find an \( x \) such that \( x' = Rx \), thus demonstrating that \( x' \) can indeed be any vector in \(\mathbb{R}^n\).
By understanding this property of invertible matrices, we can see why \( x' \) can be any vector in \(\mathbb{R}^n\).