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

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

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