Find an orthonormal basis for the subspace U of M2,2(R) spanned by −1 0 1 1 , 0 1 1 0 .
Question
Find an orthonormal basis for the subspace U of M2,2(R) spanned by
Solution
The given vectors are:
A = [ -1 0 1 1 ]
B = [ 0 1 1 0 ]
Step 1: Check if the vectors are orthogonal
Two vectors are orthogonal if their dot product is zero. In the case of matrices, we use the Frobenius inner product, which is the sum of the products of the corresponding entries.
The Frobenius inner product of A and B is:
(-10 + 01 + 11 + 10) = 1
Since the inner product is not zero, the vectors are not orthogonal.
Step 2: Gram-Schmidt Process
We can use the Gram-Schmidt process to orthogonalize the vectors and then normalize them to get an orthonormal basis.
First, let's orthogonalize:
Let u1 = A = [ -1 0 1 1 ]
To find u2, we subtract the projection of B onto A from B:
u2 = B - ((B dot u1) / (u1 dot u1)) * u1
Calculating the dot products:
B dot u1 = (0*-1 + 10 + 11 + 01) = 1 u1 dot u1 = (-1-1 + 00 + 11 + 1*1) = 3
So,
u2 = B - (1/3)*u1 = [ 0 1 1 0 ]
- (1/3)*[ -1 0
1 1 ]
= [ 1/3 1
2/3 -1/3 ]
Step 3: Normalize the vectors
To normalize, we divide each vector by its magnitude:
v1 = u1 / ||u1|| = [ -1/√3 0 1/√3 1/√3 ]
v2 = u2 / ||u2|| = [ 1/√(1/9 + 1 + 4/9 + 1/9) 1/√(1/9 + 1 + 4/9 + 1/9) 2/√(1/9 + 1 + 4/9 + 1/9) -1/3√(1/9 + 1 + 4/9 + 1/9) ]
Simplifying, we get:
v1 = [ -1/√3 0 1/√3 1/√3 ]
v2 = [ 1/√2 1/√2 2/√2 -1/6√2 ]
So, the orthonormal basis for the subspace U is {v1, v2}.
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