Collection of Maths Problems

Orthonormal bases

Task number: 2710

• Variant 1

  $\begin{pmatrix} 1 & 1 & 1 & 1 \\ 4 & 1 & 4 & 1 \\ 1 & 2 & 3 & 4 \\ \end{pmatrix}$

• Hint

  Subtract the projection to get the orthogonal vector
  $\mathbf y_i=\mathbf x_i-\sum_{j=1}^{i-1}\langle \mathbf x_i|\mathbf z_j \rangle\mathbf z_j$. Normalize it afterwards by $\mathbf z_i=\frac{\mathbf y_i}{||\mathbf y_i||}$.

• Resolution

  Normalize $\mathbf y_1=\mathbf x_1$: $||\mathbf y_1||=2$ and then $\mathbf z_1=(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T$.

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

  Normalize $\mathbf y_2$: $||\mathbf y_2||=3$ and then $\mathbf z_2=(\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2})^T$.

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

  Normalize $\mathbf y_3$: $||\mathbf y_3||=2$ and then $\mathbf z_3=(-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T$.

• Result

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

• Variant 2

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

• Resolution

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

• Result

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

• Variant 3

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

• Resolution

  $||\mathbf y_1||=3$, $\mathbf z_1=(\frac{2}{3}, 0, \frac{1}{3}, \frac{2}{3})^T$
  $\langle \mathbf x_2|\mathbf z_1 \rangle= 6$, $\mathbf y_2=(0, 3, 0, 0)^T$
  $||\mathbf y_2|| =3$, $\mathbf z_2 =(0, 1, 0, 0)^T$
  $\langle \mathbf x_3|\mathbf z_1 \rangle=9$, $\langle \mathbf x_3|\mathbf z_2 \rangle=-5$, $\mathbf y_3 =(0, 0, 0, 0)^T$,
  the third row is a linear combination of the preceding ones, we proceed with the next row $\mathbf x_3=(-4, 2, 4, 2)^T$.
  $\langle \mathbf x_3|\mathbf z_1 \rangle=0$, $\langle \mathbf x_3|\mathbf z_2 \rangle=2$, $\mathbf y_3 =(-4, 0, 4, 2)^T$
  $||\mathbf y_3||= 6$, $\mathbf z_3 =(-\frac{2}{3}, 0, \frac{2}{3}, \frac{1}{3})^T$.

• Result

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

• Variant 4

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

• Resolution

  $||\mathbf y_1||=5$, $\mathbf z_1 =(\frac{2}{5}, \frac{4}{5}, \frac{2}{5}, \frac{1}{5})^T$
  $\langle \mathbf x_2|\mathbf z_1 \rangle=-3$, $\mathbf y_2=(\frac{1}{5}, \frac{2}{5}, -\frac{4}{5}, -\frac{2}{5})^T$
  $||\mathbf y_2||=1$, $\mathbf z_2=(\frac{1}{5}, \frac{2}{5}, -\frac{4}{5}, -\frac{2}{5})^T$
  $\langle \mathbf x_3|\mathbf z_1 \rangle=4$, $\langle \mathbf x_3|\mathbf z_2 \rangle=-3$, $\mathbf y_3 =(0, 0, 0, 0)^T$,
  the third row is a linear combination of the preceding ones, we proceed with the next row $\mathbf x_3=(1, 2, 3, 4)^T$.
  $\langle \mathbf x_3|\mathbf z_1 \rangle=4$, $\langle \mathbf x_3|\mathbf z_2 \rangle=-3$, $\mathbf y_3 =(0, 0, -1, 2)^T$
  $||\mathbf y_3||=\sqrt{5}$, $\mathbf z_3 =(0, 0, -\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5})^T$

• Result

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