lectures.notes6-4 - linear independence we can do this to...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
SECTION 6.4 HOW TO GET AN ORTHOGONAL BASIS THE BASIC PROBLEM. Given a subspace W , we want to find an orthogonal basis for W . We can suppose we already have a basis for W , either because that’s how we got W in the first place, or if not, then we can produce a basis by starting with a nonzero vector in W , then adding vectors from W that aren’t in the span of the vectors we already have. So now we have a basis { u 1 , u 2 , u 3 , u 4 } for a subspace W . (Of course, 4 is not special!) Put u 1 in our proposed orthogonal basis B . The next vector in B must be orthogonal to u 1 and should surely come from u 2 . We know how to get such a vector: project u 2 onto u 1 and subtract the result from u 2 . Write this down and check out the spans. Do this again using the projection of u 3 onto the span of the first two. Write this down. CONTINUE! This is called the Gram-Schmidt Process. Since multiplying vectors by nonzero scalars does not change orthogonality or span or
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: linear independence, we can do this to make arithmetic easier. EXAMPLE. Find an orthogonal basis for the column space of -1 6 6 3-8 3 1-2 6 1-4-3 . Use the vectors in the orthogonal basis we’ve found as columns of a matrix Q . Compute Q T Q and Q T A , divide the rows of Q T A by the corresponding entries in Q T Q to get a matrix R . Finally, calculate QR . This always works and is usually done by using vectors in an orthonormal basis, rather than just an orthogonal basis. When the matrix A has linearly independent columns, then using Gram-Schmidt we can find a matrix Q whose columns are an orthonormal basis for Col A . If we put R = Q T A , then we get the QR factorization of A as A = QR , where R is an invertible upper triangular matrix. HOMEWORK: SECTION 6.4...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

lectures.notes6-4 - linear independence we can do this to...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online