Let \(M\) be an \(m \times n\) matrix, and let \(V = \{v \in \mathbb{R}^n : Mv = 0\}\) and \(W = \{M^t y : y \in \mathbb{R}^m\}\).
Prove that \(V\) and \(W\) are vector subspaces of \(\mathbb{R}^n\).
Prove that \(V = \{x \in \mathbb{R}^n : x \cdot w = 0 \text{ for all } w \in W\}\).
Solution
Part (a).
1. \(V = \{v \in \mathbb{R}^n : Mv = 0\} = \ker(M)\) is a vector subspace of \(\mathbb{R}^n\).
1.1. \(M(0_{\mathbb{R}^n}) = 0_{\mathbb{R}^m}\), so \(0_{\mathbb{R}^n} \in V\).
linear maps preserve the zero vector.
1.2. If \(u, v \in V\), then \(M(u + v) = Mu + Mv = 0 + 0 = 0\), so \(u + v \in V\).
linearity of matrix multiplication: \(M(u+v) = Mu + Mv\).
1.3. If \(v \in V\) and \(c \in \mathbb{R}\), then \(M(cv) = c(Mv) = c \cdot 0 = 0\), so \(cv \in V\).
scalar compatibility of matrix multiplication: \(M(cv) = c(Mv)\).
1.4. Hence \(V\) is a subspace of \(\mathbb{R}^n\).
subspace criterion.
2. \(W = \{M^t y : y \in \mathbb{R}^m\} = \operatorname{im}(M^t)\) is a vector subspace of \(\mathbb{R}^n\).
2.1. For \(y = 0_{\mathbb{R}^m}\), \(M^t(0_{\mathbb{R}^m}) = 0_{\mathbb{R}^n} \in W\).
\(M^t\) is linear.
2.2. If \(w_1, w_2 \in W\), then \(w_1 = M^t y_1\) and \(w_2 = M^t y_2\) for some \(y_1, y_2 \in \mathbb{R}^m\).Then \(w_1 + w_2 = M^t(y_1 + y_2) \in W\) since \(y_1 + y_2 \in \mathbb{R}^m\).
linearity of \(M^t\).
2.3. If \(w = M^t y \in W\) and \(c \in \mathbb{R}\), then \(cw = c(M^t y) = M^t(cy) \in W\) since \(cy \in \mathbb{R}^m\).
linearity of \(M^t\).
2.4. Hence \(W\) is a subspace of \(\mathbb{R}^n\).
subspace criterion.
Part (b).
3. Prove \(V \subseteq W^\perp = \{x \in \mathbb{R}^n : x \cdot w = 0 \text{ for all } w \in W\}\).
3.1. Let \(x \in V\), so \(Mx = 0\).
definition of \(V\).
3.2. Let \(w \in W\) be arbitrary, so \(w = M^t y\) for some \(y \in \mathbb{R}^m\).
definition of \(W\).
3.3. Express the dot product as matrix multiplication:\[ x \cdot w = x^t w = x^t (M^t y) = (Mx)^t y. \]
transpose identity \((AB)^t = B^t A^t\) and associativity of matrix multiplication.
3.4. Since \(Mx = 0\), \((Mx)^t y = 0^t y = 0\).
3.1.
3.5. Thus \(x \cdot w = 0\) for all \(w \in W\), so \(x \in W^\perp\).
3 and 4.
4. Prove \(W^\perp \subseteq V\).
4.1. Let \(x \in W^\perp\), so \(x \cdot w = 0\) for all \(w \in W\).
setup.
4.2. For every \(y \in \mathbb{R}^m\), the vector \(M^t y \in W\), so \(x \cdot (M^t y) = 0\).
definition of \(W\).
4.3. \(0 = x \cdot (M^t y) = x^t M^t y = (Mx)^t y = (Mx) \cdot y\) for all \(y \in \mathbb{R}^m\).
2 and transpose properties.
4.4. Choose \(y = Mx \in \mathbb{R}^m\).Then \((Mx) \cdot (Mx) = \|Mx\|^2 = 0\).
4.3.
4.5. By positive definiteness of the Euclidean norm, \(\|Mx\|^2 = 0 \implies Mx = 0\).
standard property of inner products.
4.6. Hence \(x \in V\).
definition of \(V\).
5. Conclusion: \(V = W^\perp = \{x \in \mathbb{R}^n : x \cdot w = 0 \text{ for all } w \in W\}\).
3 and 4.
Q.E.D.