Question

Problem A.4 Suppose you start out with a basis (lei), le2)..... len) that is not orthonor- mal. The Gram-Schmidt procedure is a systematic ritual for generating from it an orthonormal basis (le]). lez) en)). It goes like this: (i) Normalize the first basis vector (divide by its norm): le) lle1ll let) (ii) Find the projection of the second vector along the first, and subtract it off le2) -(eile)lei) This vector is orthogonal to lej): normalize it to get lez). (iii) Subtract from le3) its projections along lej) and lep This is orthogonal to lei) and lez); normalize it to get lej). And so on. i+(1)j+ ()k, le2) )i+(3)j + (I)k, les) (0)i (28) j+ (0) k. Use the Gram-Schmidt procedure to orthonormalize the 3-space basis le 1) (1 + i)

Answer 1

Vertical-bar e_1 rang = (1 + i) i^Hat + (1) j^Hat + (i) k^Hat Vertical-bar e_2 rang = (i) i^Hat + (3) j^Hat + (1) k^Hat Vertical-bar e_3 rang = (0) i^Hat + (28) j^Hat + (0) k^Hat (i) = > Vertical-bar e_1 rang = [(1 + i) 1 i] Therefore Vertical-bar e_1 prime rang = Vertical-bar e_1 rang/ Vertical-bar Vertical-bar e_1 Vertical-bar Vertical-bar Vertical-bar Vertical-bar e_1 Vertical-bar Vertical-bar = squareroot (1 + i)(1 - i) + 1^2 + (-i)(i) = > Vertical-bar Vertical-bar e_1 Vertical-bar Vertical-bar = squareroot 1 + 1 + 1 + 1 = squareroot 4 Therefore Vertical-bar e^1 prime rang = 1/squareroot 4 [(1 + i) 1 i] = 1/2 [(1 + i) 1 i] Vertical-bar e_2 rang = [i 3 1] Vertical-bar e_1 prime rang = 1/2 [1 + i 1 i] and lang e_1 prime Vertical-bar = 1/2 [1 - i 1 - i] = Vertical-bar e_1 prime rang^* Therefore Vertical-bar e_2 rang - lang e_1 prime Vertical-bar e_2 rang Vertical-bar e_1 prime rang = [i 3 1] - [1/2 [1 - i 1 -i]