Question 1201735: Let X be the design matrix of a linear regression problem with m rows (samples) and d columns (variables/features). Let y∈R^m be the response vector corresponding the samples in X. Recall that for some vector space V⊆R^d the orthogonal complement of V is: V^⊥:={x∈R^d∣⟨x,v⟩=0∀v∈V}
1. Prove that: Ker(X)=Ker(X^⊤X)
2. Prove that for a square matrix A:Im(A^⊤)=Ker(A)^⊥
3. Let y=Xw be a non-homogeneous system of linear equations. Assume that X is square and not invertible. Show that the system has ∞ solutions ⇔y⊥Ker(X^⊤).
4. Consider the (normal) linear system X^⊤Xw=X^⊤y. Using what you have proved above prove that the normal equations can only have a unique solution (if X^⊤X is invertible) or infinitely many solutions (otherwise).
Please help me to solve question 3 and 4.
Answer by asinus(45)   (Show Source):
You can put this solution on YOUR website! Let's tackle each part of the problem step by step.
### 1. Prove that $ \text{Ker}(X) = \text{Ker}(X^TX) $
**Proof:**
- Let $ v \in \text{Ker}(X) $. Then, by definition, $ Xv = 0 $.
- Multiplying both sides by $ X^T $, we have:
$$
X^TXv = X^T0 = 0
$$
Thus, $ v \in \text{Ker}(X^TX) $.
- Now, let $ v \in \text{Ker}(X^TX) $. Then $ X^TXv = 0 $.
- This implies $ \langle Xv, Xv \rangle = 0 $ (since $ \langle a, a \rangle = 0 $ if and only if $ a = 0 $).
- Therefore, $ Xv = 0 $, which means $ v \in \text{Ker}(X) $.
Combining both parts, we conclude:
$$
\text{Ker}(X) = \text{Ker}(X^TX)
$$
### 2. Prove that for a square matrix $ A $: $ \text{Im}(A^T) = \text{Ker}(A)^\perp $
**Proof:**
- Let $ v \in \text{Im}(A^T) $. Then there exists some $ u $ such that $ v = A^Tu $.
- For any $ w \in \text{Ker}(A) $, we have $ Aw = 0 $.
- Thus, $ \langle v, w \rangle = \langle A^Tu, w \rangle = \langle u, Aw \rangle = \langle u, 0 \rangle = 0 $.
- This shows that $ v \in \text{Ker}(A)^\perp $.
- Now, let $ v \in \text{Ker}(A)^\perp $. We need to show $ v \in \text{Im}(A^T) $.
- By the definition of orthogonal complement, $ \langle v, w \rangle = 0 $ for all $ w \in \text{Ker}(A) $.
- The rank-nullity theorem states that $ \text{dim}(\text{Im}(A)) + \text{dim}(\text{Ker}(A)) = d $.
- Since $ A $ is square, $ \text{Im}(A^T) $ has dimension equal to $ \text{dim}(\text{Ker}(A)) $.
- Therefore, $ \text{Im}(A^T) = \text{Ker}(A)^\perp $.
### 3. Show that the system $ y = Xw $ has $ \infty $ solutions $ \Leftrightarrow y \perp \text{Ker}(X^T) $
**Proof:**
- If $ y \perp \text{Ker}(X^T) $, then for any $ v \in \text{Ker}(X^T) $, we have $ \langle y, v \rangle = 0 $.
- This means that $ y $ can be expressed as $ y = Xw + v $ for some $ w $ and $ v \in \text{Ker}(X^T) $.
- Since $ v $ can take infinitely many values in $ \text{Ker}(X^T) $, there are infinitely many $ w $ that satisfy $ y = Xw $.
- Conversely, if the system has infinitely many solutions, then there exists a non-zero $ v \in \text{Ker}(X^T) $ such that $ y = Xw + v $.
- This implies $ y \perp \text{Ker}(X^T) $.
Thus, we conclude:
$$
y \perp \text{Ker}(X^T) \Leftrightarrow \text{the system has } \infty \text{ solutions}
$$
### 4. Prove that the normal equations $ X^TXw = X^Ty $ can only have a unique solution (if $ X^TX $ is invertible) or infinitely many solutions (otherwise)
**Proof:**
- If $ X^TX $ is invertible, then the normal equations have a unique solution given by:
$$
w = (X^TX)^{-1}X^Ty
$$
- If $ X^TX $ is not invertible, then $ \text{Ker}(X^TX) \neq \{0\} $. From part 1, we know:
$$
\text{Ker}(X^TX) = \text{Ker}(X)
$$
- If $ y \perp \text{Ker}(X^T) $, then the system has infinitely many solutions, as shown in part 3.
Thus, we conclude:
- The normal equations have a unique solution if $ X^TX $ is invertible.
- They have infinitely many solutions if $ X^TX $ is not invertible.
This completes the proof for all parts.
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