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Questions and Answers
Koja je formula korištena za pronalaženje matrice ortogonalne projekcije P na stupčani prostor matrice A u Problemu 1?
Koja je formula korištena za pronalaženje matrice ortogonalne projekcije P na stupčani prostor matrice A u Problemu 1?
- Formula za skalarni produkt
- Formula iz LADW-a (correct)
- Formula za vektorski produkt
- Formula za determinantu
Koji je cilj Problema 2?
Koji je cilj Problema 2?
- Pronaći linearnu jednadžbu koja najbolje odgovara skupu podataka korištenjem najmanjih kvadrata (correct)
- Pronaći matricu ortogonalne projekcije
- Pronaći kvadratnu jednadžbu koja najbolje odgovara skupu podataka korištenjem najmanjih kvadrata
- Pokazati da je ker(A) = ker(A*A
Što je potrebno da bi se tvrdilo da je A*A invertibilna u Problemu 4?
Što je potrebno da bi se tvrdilo da je A*A invertibilna u Problemu 4?
- A*A mora imati rang manji od n
- A*A mora imati rang n (correct)
- Ker(A) mora biti suprotno proporcionalan Ker(A*A
- Ker(A) mora biti jednak Ker(A*A
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Study Notes
- The text provides four problems related to linear algebra.
- Problem 1 involves finding the matrix of the orthogonal projection P onto the column space of a given matrix A.
- The projection matrix P is found using a formula from LADW.
- The size of P is not necessarily 3x2 even though the column space of A has dimension 2.
- The geometric meaning of |X - Px| is not specified.
- Problem 2 involves finding the linear equation that best fits a given set of data using least squares.
- Problem 3 involves finding the quadratic equation that best fits a given set of data using least squares.
- Problem 4 involves showing that ker(A) = ker(A*A).
- Ker(A) = ker(AA) implies that AA is invertible if and only if A has rank n.
- The invertibility of AA is important for solving the normal equation AAx - A*b.
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