Collection of Maths Problems

Orthogonal projection

Task number: 2712

• Variant 1

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

• Resolution

  We first calculate the individual projections to the vectors of the orthonormal basis \(Z=\{\mathbf z_1,\mathbf z_2,\mathbf z_3\}\):

  \(\langle \mathbf a|\mathbf z_1 \rangle=5\), \(\langle \mathbf a|\mathbf z_2 \rangle=-2\), \(\langle \mathbf a|\mathbf z_3 \rangle=1\) hence \([\mathbf p]_Z=(5, -2, 1)^T\)

  Then: \(\mathbf p=\langle \mathbf a|\mathbf z_1 \rangle\mathbf z_1+\langle \mathbf a|\mathbf z_2 \rangle\mathbf z_2+\langle \mathbf a|\mathbf z_3 \rangle\mathbf z_3\) \(=(1, 3, 2, 4)^T\)

• Result

  \([\mathbf p]_Z=(5, -2, 1)^T\), \(\mathbf p=(1, 3, 2, 4)^T\).

• Variant 2

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

• Result

  \([\mathbf p]_Z= (2, -1, \frac{7\sqrt 2}{2})^T\), \(\mathbf p=(\frac{7}{2}, 2, 1, \frac{7}{2})^T\).

• Variant 3

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

• Result

  \([\mathbf p]_Z= (5, 2, 1)^T\), \(\mathbf p=(\frac{8}{3}, 2, \frac{7}{3}, \frac{11}{3})^T\).

• Variant 4

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

• Result

  \([\mathbf p]_Z=(\frac{19}{5}, \frac{-8}{5}, \frac{9\sqrt{5}}{5})^T\), \(\mathbf p=(\frac{6}{5}, \frac{12}{5}, 1, 5)^T\).