Projectors and Projection Onto Subspaces
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The document discusses projections onto subspaces. It provides examples of projecting vectors onto lines and subspaces. For projecting a vector v onto a line defined by a vector u, it shows that the projection matrix is P=uuT/uTu. It also shows how to project vectors onto subspaces defined by matrices and how to decompose a vector into components within and orthogonal to a subspace.
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Projectors and Projection Onto Subspaces
- 1. Projector And Projection Onto Subspaces Numerical Linear Algebra Isaac Amornortey Yowetu NIMS-GHANA July 20, 2020
- 2. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Outline 1 problem of the Day 2 Projection onto a Line 3 Projection Matrix 4 Projection onto Subspace Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 3. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Problem of the Day Consider the matrix A = 1 0 0 1 1 0 What is the orthogonal projection onto the range(A) and what is the image under P of the vector (1, 2, 3)∗ ? Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 4. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Graphical Example Figure: By Nicholas Longo Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 5. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Example 1 Let v = −2 3 and u = −1 1 What is the orthogonal projector P onto range(u), and what is the image under P of the vector (v) ? Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 6. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Solution Projection Matrix Here we would like to find a projection matrix P: P = P2 and PT = P P = uuT uT u proju(v) = Pv = uuT uT u v Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 7. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Solution Continue... P = uuT uT u u · uT = −1 1 (−1 1) = 1 −1 −1 1 uT · u = (−1 1) −1 1 = 2 ∴ P = 1 2 1 −1 −1 1 = 0.5 −0.5 −0.5 0.5 Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 8. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace proju(v) = Pv = 0.5 −0.5 −0.5 0.5 −2 3 = −2.5 2.5 Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 9. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Problem of the Day Consider the matrix A = 1 0 0 1 1 0 What is the orthogonal projection onto the range(A) and what is the image under P of the vector (1, 2, 3)∗ ? Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 10. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Solution Projection Matrix Here we would like to find a projection matrix P: P = P2 and PT = P P = uuT uT u But considering u to be matrix A, then: projA(v) = Pv = AAT AT A v Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 11. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Solution Continue... Remark But we don’t divide matrices. Hence, we will then have inverse instead of matrix division. We can consider multiplying the matrix A by the pseudo-inverse to get matrix P. Pseudo-inverse = (AT A)−1 AT Then, P = A(AT A)−1 AT Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 12. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Solution Continue... AT A = 1 0 1 0 1 0 1 0 0 1 1 0 = 2 0 0 1 (AT A)−1 = 0.5 0 0 1 (AT A)−1 AT = 0.5 0 0 1 1 0 1 0 1 0 = 0.5 0 0.5 0 1 0 A(AT A)−1 AT = 1 0 0 1 1 0 0.5 0 0.5 0 1 0 = 0.5 0 0.5 0 1 0 0.5 0 0.5 Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 13. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Considering P = 0.5 0 0.5 0 1 0 0.5 0 0.5 and 1 2 3 Then, Pv = 0.5 0 0.5 0 1 0 0.5 0 0.5 1 2 3 = 2 2 2 Conclusion: P is our orthogonal Projection onto Range(A) and (2, 2, 2)T is the image under P of the vector(v). Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 14. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Example 2 Let A be the 2-dimensional subspace of R3 spanned by the orthogonal vectors u1 = (1, 0, 1) and u2 = (0, 1, 0). Write the vector v = (1, 2, 3) as the sum of vector in A and a vector orthogonal to A. Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 15. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Solution Let x = proju1 v + proju2 v (1) x = λu1 + βu2 (2) x = u1 · v u1 · u1 u1 + u2 · v u2 · u2 u2 (3) x = 1 0 1 · 1 2 3 1 0 1 · 1 0 1 1 0 1 + 0 1 0 · 1 2 3 0 1 0 · 0 1 0 0 1 0 (4) Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces
- 16. problem of the Day Projection onto a Line Projection Matrix Projection onto Subspace Solution Continue... x = 2 1 0 1 + 2 0 1 0 = 2 2 2 (5) v ⊥ A = 1 2 3 − 2 2 2 = −1 0 1 (6) Conclusion: Vector x is our sum of vector v in A and (v ⊥ A) is vector orthogonal to A. Isaac Amornortey Yowetu NIMS-GHANA Projector And Projection Onto Subspaces