Question

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 \(A \mathbf{z} \in W^{\perp} .\)

Answer 1

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 $A\mathbf{z} \in W^{\perp}$.