MAT188H1 Midterm: MAT188H1_20169_641483478668mat188tut11sol
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MAT188H1 Full Course Notes
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Faculty of applied science & engineering, university of toronto. 1 + + a2 n. (c) let {v1, v2, v3} be an orthonormal basis for r3. + (a2v2) (b1v1) + (a2v2) (b2v2) + . + (anvn) (b1v1) + (anvn) (b2v2) + . , vn} form an orthonormal set, by de nition we have vi vj =(cid:26)1, Hence, using the third property above, (aivi) (bjvj) = aibj(vi vj) =(cid:26)aibj, 0 if i = j, if i 6= j. Therefore most terms in our expression for x y are zero, except those where both indices are equal, leaving x y = a1b1 + a2b2 + . + anbn (b) recall that ||x||2 = x x. Applying part (a) with y = x, we get. 1 of 7 (c) by part (a), x v1 = a1, x v2 = a2. So x v1 = 4 implies that a1 = 4.
