Sbírka matematických úloh

Ortonormální báze

Úloha číslo: 2699

• Varianta 1

  \( \begin{pmatrix} 1 & 1 & 1 & 1 \\ 4 & 1 & 4 & 1 \\ 1 & 2 & 3 & 4 \\ \end{pmatrix} \)

• Nápověda

  Odečtením projekcí spočteme nejprve kolmý vektor
  \(\mathbf y_i=\mathbf x_i-\sum_{j=1}^{i-1}\langle \mathbf x_i|\mathbf z_j \rangle\mathbf z_j\); a ten pak normalizujeme \(\mathbf z_i=\frac{\mathbf y_i}{||\mathbf y_i||}\).

• Řešení

  Normalizujeme \(\mathbf y_1=\mathbf x_1\): \(||\mathbf y_1||=2\) a odtud \(\mathbf z_1=(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T\).

  \(\langle \mathbf x_2|\mathbf z_1 \rangle = 5\) a proto \(\mathbf y_2=(\frac{3}{2}, -\frac{3}{2}, \frac{3}{2}, -\frac{3}{2})^T\).

  Normalizujeme \(\mathbf y_2\): \(||\mathbf y_2||=3\) a dostaneme \(\mathbf z_2=(\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2})^T\).

  \(\langle \mathbf x_3|\mathbf z_1 \rangle = 5\), \(\langle \mathbf x_3|\mathbf z_2 \rangle = -1\) a tedy \(\mathbf y_3=(-1, -1, 1, 1)^T\).

  Normalizujeme \(\mathbf y_3\): \(||\mathbf y_3||=2\) a máme \(\mathbf z_3=(-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T\).

• Výsledek

  \(Z=\{(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T, (\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2})^T, (-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T\}\).

• Varianta 2

  \( \begin{pmatrix} 0 & 3 & 4 & 0 \\ 0 & 0 & 5 & 0 \\ 2 & 1 & 0 & 2 \\ \end{pmatrix} \)

• Řešení

  \(||\mathbf y_1||= 5 \), \(\mathbf z_1 =(0, \frac{3}{5}, \frac{4}{5}, 0)^T\)
  \(\langle \mathbf x_2|\mathbf z_1 \rangle= 4\), \(\mathbf y_2=(0, -\frac{12}{5}, \frac{9}{5}, 0)^T\)
  \(||\mathbf y_2||=3\), \(\mathbf z_2 =(0, -\frac{4}{5}, \frac{3}{5}, 0)^T\)
  \(\langle \mathbf x_3|\mathbf z_1 \rangle=\frac{3}{5}\), \(\langle \mathbf x_3|\mathbf z_2 \rangle=-\frac{4}{5}\), \(\mathbf y_3 =(2, 0, 0, 2)^T\)
  \(||\mathbf y_3||=2\sqrt{2}\), \(\mathbf z_3 =(\frac{\sqrt{2}}{2}, 0, 0, \frac{\sqrt{2}}{2})^T\)

• Výsledek

  \(Z=\{(0, \frac{3}{5}, \frac{4}{5}, 0)^T, (0, -\frac{4}{5}, \frac{3}{5}, 0)^T, (\frac{\sqrt{2}}{2}, 0, 0, \frac{\sqrt{2}}{2})^T\}\).

• Varianta 3

  \( \begin{pmatrix} 2 & 0 & 1 & 2 \\ 4 & 3 & 2 & 4 \\ 6 & -5 & 3 & 6 \\ -4 & 2 & 4 & 2 \\ \end{pmatrix} \)

• Řešení

  \(||\mathbf y_1||=3\), \(\mathbf z_1=(\frac{2}{3}, 0, \frac{1}{3}, \frac{2}{3})^T\)
  \(\langle \mathbf x_2|\mathbf z_1 \rangle= 6\), \(\mathbf y_2=(0, 3, 0, 0)^T\)
  \(||\mathbf y_2|| =3\), \(\mathbf z_2 =(0, 1, 0, 0)^T\)
  \(\langle \mathbf x_3|\mathbf z_1 \rangle=9\), \(\langle \mathbf x_3|\mathbf z_2 \rangle=-5\), \(\mathbf y_3 =(0, 0, 0, 0)^T\),
  třetí řádek je kombinací předchozích, volíme tedy následující řádek \(\mathbf x_3=(-4, 2, 4, 2)^T\).
  \(\langle \mathbf x_3|\mathbf z_1 \rangle=0\), \(\langle \mathbf x_3|\mathbf z_2 \rangle=2\), \(\mathbf y_3 =(-4, 0, 4, 2)^T\)
  \(||\mathbf y_3||= 6\), \(\mathbf z_3 =(-\frac{2}{3}, 0, \frac{2}{3}, \frac{1}{3})^T\).

• Výsledek

  \(Z=\{(\frac{2}{3}, 0, \frac{1}{3}, \frac{2}{3})^T, (0, 1, 0, 0)^T, (-\frac{2}{3}, 0, \frac{2}{3}, \frac{1}{3})^T\}\).

• Varianta 4

  \( \begin{pmatrix} 2 & 4 & 2 & 1 \\ -1 & -2 & -2 & -1 \\ 1 & 2 & 4 & 2 \\ 1 & 2 & 3 & 4 \\ \end{pmatrix} \)

• Řešení

  \(||\mathbf y_1||=5\), \(\mathbf z_1 =(\frac{2}{5}, \frac{4}{5}, \frac{2}{5}, \frac{1}{5})^T\)
  \(\langle \mathbf x_2|\mathbf z_1 \rangle=-3\), \(\mathbf y_2=(\frac{1}{5}, \frac{2}{5}, -\frac{4}{5}, -\frac{2}{5})^T\)
  \(||\mathbf y_2||=1\), \(\mathbf z_2=(\frac{1}{5}, \frac{2}{5}, -\frac{4}{5}, -\frac{2}{5})^T\)
  \(\langle \mathbf x_3|\mathbf z_1 \rangle=4\), \(\langle \mathbf x_3|\mathbf z_2 \rangle=-3\), \(\mathbf y_3 =(0, 0, 0, 0)^T\),
  třetí řádek je kombinací předchozích, volíme tedy následující řádek \(\mathbf x_3=(1, 2, 3, 4)^T\).
  \(\langle \mathbf x_3|\mathbf z_1 \rangle=4\), \(\langle \mathbf x_3|\mathbf z_2 \rangle=-3\), \(\mathbf y_3 =(0, 0, -1, 2)^T\)
  \(||\mathbf y_3||=\sqrt{5}\), \(\mathbf z_3 =(0, 0, -\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5})^T\)

• Výsledek

  \(Z =\{(\frac{2}{5}, \frac{4}{5}, \frac{2}{5}, \frac{1}{5})^T, (\frac{1}{5}, \frac{2}{5}, -\frac{4}{5}, -\frac{2}{5})^T, (0, 0, -\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5})^T\}\).