thanks Prove that any orthogonal set of three vectors in Rn is linearly independ

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Prove that any orthogonal set of three vectors in Rn is linearly independent. [Hint: Apply the reasoning used in the proof of Theorem 4 to the dependency equation.] Let B = {P1, P2, ,Pn} be an ordered orthogonal basis for Rn and let X Rn. Then X = x'1P1 + x'2P2 + +x'nPn where x'i = X.Pi/Pi. Pi In particular, the B coordinate vector for X is X' = [x'1,x'2, ,x'n], where the x'i are given by formula (6.12). Proof. Since B is a basis, we may expand X as in equation (6.11). We take the dot product of both sides of this equation with Pj and use Pi. Pj = 0, i j, finding X.Pi = 0+0+ +x'iPi. Pi+0+ +0 = x'iPi. Pi which is equivalent to our theorem. Notice that in Example 4 it was given that B forms a basis of R3. We, of course, could have checked this by showing that B is linearly independent. (We need to show only linear independence, since R3 is three-dimensional.) However, since the Pi, are mutually perpendicular, it seems apparent that they are linearly independent - how could, say, P3 be a linear combination of P1 and P2 if it is perpendicular to the plane they span? (See Figure 6.4.) The following theorem is proved using much the same argument as in the proof of Theorem 4. The proof is indicated in Exercise 17.

Explanation / Answer

Sorry, but chrome is bugging out, cannot use the nice formatting. :( Let x y and z be orth vectors. Suppose a,b,c are scalars such that ax + by + cz = 0 Now dot both sides of this equation by x: (* is dot prod) (ax + by + cz)*x = 0*x Now distribute the dot prod on left. ax*x + by*x + cz*x = 0 Now use x*x = |x|^2 and the fact that x,y,z are orth. a|x|^2 + 0 + 0 = 0 a|x|^2 = 0 Since an orth set has no 0 vectors, this means a=0. Now mult orig equation on right by y. Gives: ax*y + by*y + cz*y = 0*y b|y|^2 = 0 b =0 Similarly for z: ax*z + by*z + cz*z =0*z c|z|^2 = 0 c = 0 Therefore a=b=c=0 ----> x,y,z are linearly independent.

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