Question 1160970: Let v=[4, -9, 1, 9]. Find a basis of the subspace of R^4 consisting of all vectors perpendicular to v.
Found 2 solutions by rothauserc, ikleyn:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! v = [4, -9, 1, 9]
Let u be a vector in R^4 and let R^4 be the set of 4 by 1 column vectors
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Then let W = { u an element of R^4 such that vu = 0 }
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Show subspace requirements are satisfied, that is, zero vector in R^4 is in W and W is closed under addition and scaler multiplication.
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The zero element in R^4 is 0, the 4 by 1 column vector whose entries are all 0, then v0 = 0, therefore 0 is an element in W
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Suppose u, w are elements of W and c is an element of R, then vu = vw = 0 and
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v(u +w) = vu +vw = 0, therefore u +w is an element in W
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Since vu = 0, v(cu) = cvu = c0 = 0
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Vectors u, w in R^4 are said to be perpendicular if u^Tw = 0
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let v = u^T and we have show the set of vectors perpendicular to any given vector is a subspace of R^4
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Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
The other tutor retold the general theory, but the meaning of the problem and the meaning of the request
is to point (to find) three concrete linearly independent vectors in  perpendicular to vector V.
It was not done in the post by the other tutor, so I will do it right now.
The first vector in , perpendicular to vector v, is x = (1,0,-4,0).
Indeed, you can check it on your own that the scalar product of vectors v and x is equal to zero
(v,x) = 4*1 + (-9)*0 + 1*(-4) + 9*0 = 4 + 0 - 4 + 0 = 0.
The second vector in , perpendicular to vector v, is y = (0,-1,-9,0).
Indeed, you can check it on your own that the scalar product of vectors v and y is equal to zero
(v,y) = 4*0 + (-9)*(-1) + 1*(-9) + 9*0 = 0 + 9 - 9 + 0 = 0.
Finally, the third vector in , perpendicular to vector v, is z = (0,0,-9,1).
Indeed, you can check it on your own that the scalar product of vectors v and z is equal to zero
(v,z) = 4*0 + (-9)*0 + 1*(-9) + 9*1 = 0 + 0 - 9 + 9 = 0.
Next, it is OBVIOUS that the three vectors
x = (1, 0, -4, 0),
y = (0,-1, -9, 0) and
z = (0, 0, -9, 1)
are linearly independent (due to construction of their components).
Thus we constructed (guessed, based on intuition) three linearly independent vectors in perpendicular to the given vector v.
Hence, these three vectors x, y and z form a basis in the orthogonal complement to vector V in .
It is what has to be done.
Solved, explained and completed.
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