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Let $$A$$ be an $$n \times n$$ real self-adjoint matrix and $$\mathbf{v}$$ an eigenvector with eigenvalue $$\lambda$$. Let $$W=\operatorname{span}\{\mathbf{v}\}$$.

a) If $$\mathbf{w} \in W$$, show that $$A \mathbf{w} \in W$$
b) If $$\mathbf{z} \in W^{\perp}$$, show that $$A \mathbf{z} \in W^{\perp} .$$
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a) Let $\mathbf{w} \in W$, i.e., $\mathbf{w} = c\mathbf{v}$ for some scalar $c$. Then, $$A\mathbf{w} = A(c\mathbf{v}) = c(A\mathbf{v}) = c(\lambda \mathbf{v}) = \lambda(c\mathbf{v}) = \lambda \mathbf{w}$$. Thus, $A\mathbf{w} \in W$.

b) Let $\mathbf{z} \in W^{\perp}$. Then, $\mathbf{z}^{\top} \mathbf{v} = \mathbf{0}$. Taking the transpose of both sides, we have $\mathbf{v}^{\top} \mathbf{z} = \mathbf{0}$. Therefore, $$(A\mathbf{z})^{\top} \mathbf{v} = \mathbf{z}^{\top} A^{\top} \mathbf{v} = \mathbf{z}^{\top} A \mathbf{v} = \mathbf{z}^{\top} \lambda \mathbf{v} = \lambda \mathbf{z}^{\top} \mathbf{v} = \mathbf{0}$$. Since $\mathbf{v} \neq \mathbf{0}$, it follows that $A\mathbf{z} \in W^{\perp}$.
by Diamond (55.6k points)

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