【线性代数】6.4 exercise20

answer

To show that A" role="presentation">A$A$ and Q" role="presentation">Q$Q$ have the same column space, we'll use the given hints:

### Step 1: Show that Col A⊆Col Q" role="presentation">Col A⊆Col Q$\text{Col }A\subseteq \text{Col }Q$
Given y∈Col A" role="presentation">y∈Col A$y\in \text{Col }A$, we can write y=Ax" role="presentation">y=Ax$y=Ax$ for some vector x" role="presentation">x$x$.

Since A=QR" role="presentation">A=QR$A=QR$ and R" role="presentation">R$R$ is invertible, we have:

Let x′=Rx" role="presentation">x′=Rx$x^{\prime }=Rx$. Since R" role="presentation">R$R$ is invertible, x′" role="presentation">x′$x^{\prime }$ can be any vector in Rn" role="presentation">Rn${\mathbb{R}}^n$. Thus:

This shows that y" role="presentation">y$y$ is also in the column space of Q" role="presentation">Q$Q$, i.e., Col A⊆Col Q" role="presentation">Col A⊆Col Q$\text{Col }A\subseteq \text{Col }Q$.

### Step 2: Show that Col Q⊆Col A" role="presentation">Col Q⊆Col A$\text{Col }Q\subseteq \text{Col }A$
Given y∈Col Q" role="presentation">y∈Col Q$y\in \text{Col }Q$, we can write y=Qx" role="presentation">y=Qx$y=Qx$ for some vector x" role="presentation">x$x$.

Since A=QR" role="presentation">A=QR$A=QR$ and R" role="presentation">R$R$ is invertible, we can multiply both sides by R−1" role="presentation">R−1$R^{-1}$:
\[ Q = AR^{-1} \]

Thus:

Let x′=R−1x" role="presentation">x′=R−1x$x^{\prime }=R^{-1}x$. Since R" role="presentation">R$R$ is invertible, x′" role="presentation">x′$x^{\prime }$ can be any vector in Rn" role="presentation">Rn${\mathbb{R}}^n$. Thus:

This shows that y" role="presentation">y$y$ is also in the column space of A" role="presentation">A$A$, i.e., Col Q⊆Col A" role="presentation">Col Q⊆Col A$\text{Col }Q\subseteq \text{Col }A$.

### Conclusion
Since we've shown that Col A⊆Col Q" role="presentation">Col A⊆Col Q$\text{Col }A\subseteq \text{Col }Q$ and Col Q⊆Col A" role="presentation">Col Q⊆Col A$\text{Col }Q\subseteq \text{Col }A$, we conclude that Col A=Col Q" role="presentation">Col A=Col Q$\text{Col }A=\text{Col }Q$.

Therefore, A" role="presentation">A$A$ and Q" role="presentation">Q$Q$ have the same column space.

details

To understand why x′" role="presentation">x′$x^{\prime }$ can be any vector in Rn" role="presentation">Rn${\mathbb{R}}^n$, let's break it down step by step.

### Understanding the Concept

Given that R" role="presentation">R$R$ is an invertible matrix:

1. **Invertibility of R" role="presentation">R$R$**: Since R" role="presentation">R$R$ is invertible, there exists a matrix R−1" role="presentation">R−1$R^{-1}$ such that:
   

   where I" role="presentation">I$I$ is the identity matrix.

2. **Transformation by R" role="presentation">R$R$**: Any vector x∈Rn" role="presentation">x∈Rn$x\in {\mathbb{R}}^n$ can be transformed by multiplying it by R" role="presentation">R$R$, resulting in a new vector x′" role="presentation">x′$x^{\prime }$. We write:
   

3. **Invertible Transformation**: Because R" role="presentation">R$R$ is invertible, we can always find x" role="presentation">x$x$ given x′" role="presentation">x′$x^{\prime }$ by multiplying x′" role="presentation">x′$x^{\prime }$ by R−1" role="presentation">R−1$R^{-1}$:
   

### Why x′" role="presentation">x′$x^{\prime }$ Can Be Any Vector in Rn" role="presentation">Rn${\mathbb{R}}^n$

Let's see why x′" role="presentation">x′$x^{\prime }$ can be any vector in Rn" role="presentation">Rn${\mathbb{R}}^n$:

1. **Surjectivity**: The function f:Rn→Rn" role="presentation">f:Rn→Rn$f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ defined by f(x)=Rx" role="presentation">f(x)=Rx$f(x)=Rx$ is surjective because for any vector x′" role="presentation">x′$x^{\prime }$ in the output space (the codomain), there exists an x" role="presentation">x$x$ in the input space (the domain) such that Rx=x′" role="presentation">Rx=x′$Rx=x^{\prime }$. This is because R" role="presentation">R$R$ has full rank (since it's invertible).

2. **Vector Space Mapping**: Since R" role="presentation">R$R$ maps Rn" role="presentation">Rn${\mathbb{R}}^n$ to itself in a bijective manner (one-to-one and onto), for every vector x′∈Rn" role="presentation">x′∈Rn$x^{\prime }\in {\mathbb{R}}^n$, there exists a vector x∈Rn" role="presentation">x∈Rn$x\in {\mathbb{R}}^n$ such that x′=Rx" role="presentation">x′=Rx$x^{\prime }=Rx$.

3. **Existence of x" role="presentation">x$x$**: Given any vector x′∈Rn" role="presentation">x′∈Rn$x^{\prime }\in {\mathbb{R}}^n$, we can always find a corresponding vector x" role="presentation">x$x$ using x=R−1x′" role="presentation">x=R−1x′$x=R^{-1}{x}^{\prime }$. Therefore, x′" role="presentation">x′$x^{\prime }$ can be any vector in Rn" role="presentation">Rn${\mathbb{R}}^n$.

### Example

Let's consider a concrete example with R" role="presentation">R$R$:

1. Suppose R" role="presentation">R$R$ is a 2x2 invertible matrix:
   

2. We can find R−1" role="presentation">R−1$R^{-1}$:
   

3. Now, given any vector x′∈R2" role="presentation">x′∈R2$x^{\prime }\in {\mathbb{R}}^2$, say x′=(45)" role="presentation">x′=(45)$x^{\prime }=\left(\begin{array}{c}4\\ 5\end{array}\right)$:
   

This shows that for any x′" role="presentation">x′$x^{\prime }$, we can find an x" role="presentation">x$x$ such that x′=Rx" role="presentation">x′=Rx$x^{\prime }=Rx$, thus demonstrating that x′" role="presentation">x′$x^{\prime }$ can indeed be any vector in Rn" role="presentation">Rn${\mathbb{R}}^n$.

By understanding this property of invertible matrices, we can see why x′" role="presentation">x′$x^{\prime }$ can be any vector in Rn" role="presentation">Rn${\mathbb{R}}^n$.