Exercise 1.8.1

Answers

We have

\begin{array}{l} x^{T} x & = {(c_{1} v_{1} + \dots + c_{n} v_{n})}^{T} (c_{1} v_{1} + \dots + c_{n} v_{n}) \\ = (c_{1} v_{1}^{T} + \dots + c_{n} v_{n}^{T}) (c_{1} v_{1} + \dots + c_{n} v_{n}) \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{n} c_{i} v_{i}^{T} c_{j} v_{j} \\ = \sum_{i = 1}^{n} c_{i} c_{i} v_{i}^{T} v_{i} \\ = c_{1}^{2} + \dots + c_{n}^{2} \end{array}

and

\begin{array}{l} x^{T} Sx & = x^{T} S (c_{1} v_{1} + \dots + c_{n} v_{n}) \\ = x^{T} (c_{1} S v_{1} + \dots + c_{n} S v_{n}) \\ = x^{T} (c_{1} λ_{1} v_{1} + \dots + c_{n} λ_{n} v_{n}) \\ = {(c_{1} v_{1} + \dots + c_{n} v_{n})}^{T} (c_{1} λ_{1} v_{1} + \dots + c_{n} λ_{n} v_{n}) \\ = λ_{1} c_{1}^{2} + \dots + λ_{n} c_{n}^{2} \end{array}

If we take the symmetric matrix as $I$ , which has eigenvalues of $1$ , we can recover the first equation from the second one.

niuers

2020-03-20 00:00