Algebra lineare - anno 2013-14

Per ognuna delle seguenti matrici reali simmetriche $A$ trovare una matrice ortogonale $M$ tale che $\phantom{.}^tMAM$ sia diagonale (possibili risposte a lato): $$A=\begin{pmatrix} 1&1&-1\\ 1&1&2\\ -1&2&1 \end{pmatrix} \qquad\text{risposta: } M=\begin{pmatrix} 0&\frac{1+\sqrt{3}}{\sqrt{2}\sqrt{3+\sqrt{3}}}&\frac{1-\sqrt{3}}{\sqrt{2}\sqrt{3+\sqrt{3}}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}\sqrt{3+\sqrt{3}}}&\frac{1}{\sqrt{2}\sqrt{3-\sqrt{3}}}\\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}\sqrt{3+\sqrt{3}}}&\frac{-1}{\sqrt{2}\sqrt{3-\sqrt{3}}} \end{pmatrix} \ ,\quad \phantom{.}^tMAM=\begin{pmatrix} 3&0&0\\ 0&\sqrt{3}&0\\ 0&0&-\sqrt{3} \end{pmatrix}$$ $$A=\begin{pmatrix} 0&-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}&\frac{1}{2}&\frac{1}{2}\\ \frac{1}{\sqrt{2}}&\frac{1}{2}&\frac{1}{2} \end{pmatrix} \qquad\text{risposta: } M=\begin{pmatrix} 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{-1}{2}&\frac{1}{2}\\ \frac{1}{\sqrt{2}}&\frac{1}{2}&\frac{-1}{2} \end{pmatrix} \ ,\quad \phantom{.}^tMAM=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&-1 \end{pmatrix}$$ $$A=\begin{pmatrix} -1&\frac{3}{2}&\frac{1}{2}\sqrt {3}\\ \frac{3}{2}&\frac{1}{4}&\frac{3}{4}\sqrt {3}\\ \frac{1}{2}\sqrt{3}&\frac{3}{4}\sqrt{3}&\frac{-5}{4} \end{pmatrix} \qquad\text{risposta: } M=\begin{pmatrix} \frac{1}{2}&\frac{3}{\sqrt{13}}&\frac{\sqrt{3}}{2\sqrt{13}}\\ \frac{3}{4}&\frac{-2}{\sqrt{13}}&\frac{3\sqrt{3}}{4\sqrt{13}}\\ \frac{\sqrt{3}}{4}&0&\frac{-\sqrt{13}}{4} \end{pmatrix} \ ,\quad \phantom{.}^tMAM=\begin{pmatrix} 2&0&0\\ 0&-2&0\\ 0&0&-2 \end{pmatrix}$$