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Exercise 7.D.6

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Proof. We will use the orthonormal basis $\sqrt{\frac{1}{2}}, \sqrt{\frac{3}{2}} x, \sqrt{\frac{45}{8}} (x^{2} - \frac{1}{3})$ of $P_{2} (ℝ)$ , which was found in the example.

We have

\begin{aligned} D \sqrt{\frac{1}{2}} & = 0 \\ D \sqrt{\frac{3}{2}} x & = \sqrt{3} (\sqrt{\frac{1}{2}}) \\ D \sqrt{\frac{45}{8}} (x^{2} - \frac{1}{3}) & = \sqrt{15} (\sqrt{\frac{3}{2}} x) . \end{aligned}

Therefore

M (D) = (\begin{matrix} 0 & \sqrt{3} & 0 \\ 0 & 0 & \sqrt{15} \\ 0 & 0 & 0 \end{matrix}) .

Thus

M (D^{∗} D) = (\begin{matrix} 0 & 0 & 0 \\ \sqrt{3} & 0 & 0 \\ 0 & \sqrt{15} & 0 \end{matrix}) (\begin{matrix} 0 & \sqrt{3} & 0 \\ 0 & 0 & \sqrt{15} \\ 0 & 0 & 0 \end{matrix}) = (\begin{matrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 15 \end{matrix}) .

Thus, the singular values of $D$ are $0, \sqrt{3}, \sqrt{15}$ . □

celiopassos

2017-10-06 00:00

Exercise 7.D.6

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