Find the basis for the following vector space. Please state the dimension of the vector space. S consists of all x in R3 such that x is orthogonal to n=(2,3,2)
The dimension of vector space R3 is 3.
n=(2,3,2) occupies one of the three dimensions, so the subspace orthogonal to n has a dimension of two, i.e. two vectors span the remaining subspace.
The basis of the remaining subspace can be in many forms. One way is to start with an arbitrary vector, say a1=(1,1,0) and apply Gram-Schmidt process to transform it into a vector orthogonal to n. After normalization, we get A1=(7,2,-10)/(3sqrt(17)).
Note that A1.n=0.
Similarly, if we start with a2=(1,0,1), and apply Gram-Schmidt process, we get A2=(1,0,-1)/sqrt(2).
Note also that A2.n=0, and A1.A2=0.
Thus we have a basis for R3 {n,A1,A2} of which two are orthogonal to n. (n being one of the three vectors of the basis).
... and the basis for the subspace orthogonal to n is {A1,A2}.
To find the basis for the vector space S, we need to determine the set of vectors that are orthogonal to n = (2, 3, 2).
Let's denote a vector in the vector space S as x = (x1, x2, x3). For x to be orthogonal to n, the dot product of x and n must be zero:
x · n = 0
Substituting the values of n and x, we have:
(2x1 + 3x2 + 2x3) = 0
Now, let's solve this equation to find the basis for S.
First, let's express x3 in terms of x1 and x2:
2x3 = -(2x1 + 3x2)
x3 = -(2/2)x1 - (3/2)x2
x3 = -x1 - (3/2)x2
So, any vector x in S can be written as:
x = (x1, x2, -x1 - (3/2)x2)
To find the basis for S, we need to determine the linearly independent vectors that can span the entire vector space. Based on the above equation, we can choose two vectors that satisfy the conditions.
Let's choose two specific vectors for x:
1. x1 = 1, x2 = 0:
x1 = 1, x2 = 0, x3 = -1
2. x1 = 0, x2 = 1:
x1 = 0, x2 = 1, x3 = -(3/2)
Therefore, the basis for the vector space S is:
{(1, 0, -1), (0, 1, -(3/2))}
The dimension of the vector space is equal to the number of vectors in its basis. In this case, the dimension of S is 2.