Collection of Maths Problems

Extension of the basis

Task number: 2711

• Variant 1

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

• Hint

  We first extend the basis of the row space to a basis of \(\mathbb R^4\): We transform the matrix to the echelon form. Then we extend the basis by adding rows with "missing" pivots.

  Note also that the kernel is the orthogonal complement of the row space, i.e. the orthogonal extension of the basis is also a basis of the kernel.

• Resolution

  We first transform the matrix to the echelon form: \( \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 2 & 2 \\ \end{pmatrix} \).

  Now we may choose e.g. \(\mathbf x_4=(0, 0, 0, 1)^T\). When we orthonormalize the added vector, it wil yield a basis of the kernel of the matrix.

  Extension of the basis \(Z=\{(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T, (\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{2})^T, (-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2}, \frac{1}{2})^T\}\).

  \(\langle \mathbf x_4|\mathbf z_1 \rangle=\frac{1}{2}\), \(\langle \mathbf x_4|\mathbf z_2 \rangle=-\frac{1}{2}\), \(\langle \mathbf x_4|\mathbf z_3 \rangle=\frac{1}{2}\), a tudíž \(\mathbf y_4=(\frac{1}{4}, -\frac{1}{4}, -\frac{1}{4}, \frac{1}{4})^T\),

  Normalize \(\mathbf y_4\): \(||\mathbf y_4||=\frac{1}{2}\) and get \(\mathbf z_4=(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \frac{1}{2})^T\).

• Result

  \(\mathbf z_4=(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \frac{1}{2})^T\).

• Variant 2

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

• Resolution

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

  \(\mathbf x_4=(0, 0, 0, 1)^T\)

  \(\langle \mathbf x_4|\mathbf z_1 \rangle= 0\), \(\langle \mathbf x_4|\mathbf z_2 \rangle= 0\), \(\langle \mathbf x_4|\mathbf z_3 \rangle=\frac{\sqrt{2}}{2}\), \(\mathbf y_4=(-\frac{1}{2}, 0, 0, \frac{1}{2})^T\) \(||\mathbf y_4||=\frac{\sqrt{2}}{2}\), \(\mathbf z_4=(-\frac{\sqrt{2}}{2}, 0, 0, \frac{\sqrt{2}}{2})^T\).

• Result

  \(\mathbf z_4=(-\frac{\sqrt{2}}{2}, 0, 0, \frac{\sqrt{2}}{2})^T\).

• Variant 3

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

• Resolution

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

  \(\mathbf x_4=(0, 0, 0, 1)^T\),

  \(\langle \mathbf x_4|\mathbf z_1 \rangle=\frac{2}{3}\), \(\langle \mathbf x_4|\mathbf z_2 \rangle=0\), \(\langle \mathbf x_4|\mathbf z_3 \rangle=\frac{1}{3}\), \(\mathbf y_4=(-\frac{2}{9}, 0, -\frac{4}{9}, \frac{4}{9})^T\), \(||\mathbf y_4||=\frac{2}{3}\), \(\mathbf z_4=(-\frac{1}{3}, 0, -\frac{2}{3}, \frac{2}{3})^T\).

• Result

  \(\mathbf z_4=(-\frac{1}{3}, 0, -\frac{2}{3}, \frac{2}{3})^T\).

• Variant 4

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

• Resolution

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

  \(\mathbf x_4=(0, 1, 0, 0)^T\),

  \(\langle \mathbf x_4|\mathbf z_1 \rangle=\frac{4}{5}\), \(\langle \mathbf x_4|\mathbf z_2 \rangle=\frac{2}{5}\). \(\langle \mathbf x_4|\mathbf z_3 \rangle= 0\), \(\mathbf y_4=(-\frac{2}{5}, \frac{1}{5}, 0, 0)^T\),

  \(||\mathbf y_4||=\frac{\sqrt{5}}{5}\), \(\mathbf z_4=(-\frac{2\sqrt{5}}{5}, \frac{\sqrt{5}}{5}, 0, 0)^T\).

• Result

  \(\mathbf z_4=(-\frac{2\sqrt{5}}{5}, \frac{\sqrt{5}}{5}, 0, 0)^T\).