Exercise 1.7.1

Answers

$y^{T} Sx = y^{T} λx = λ y^{T} x$
$x^{T} Sy = x^{T} αy = α x^{T} y$
Since $y^{T} Sx$ is real number, it equals to its transpose, i.e. $y^{T} Sx = x^{T} S^{T} y = x^{T} Sy$ . So we have

\begin{array}{l} λ y^{T} x & = α x^{T} y \\ λ y^{T} x - α x^{T} y & = 0 \\ λ y^{T} - α y^{T} x & = 0 \\ (λ - α) y^{T} x & = 0 \end{array}

So if $λ \neq α$ , we have $y^{T} x = 0$ , i.e. $y$ and $x$ are orthogonal eigenvectors.

Note we used the fact that $y^{T} x$ is a real number, so we have $y^{T} x = {(y^{T} x)}^{T} = x^{T} y$

niuers

2020-03-20 00:00